the Base BC Let there be imagined another Triangle DEF having an Angle D equal to the Angle A and the Sides DE DF equal to AB AC Now since the Sides AB AC are equal the Four Lines AB AC DE DF shall be equal Demonstration because the Sides AB DE AC DF are equal as also the Angles A and D if we put the Triangle DEF on the Triangle ABC the Line DE shall fall upon AB and DF on AC and FE on BC by the Fourth therefore the Angle DEF shall be equal to ABC and since one part of the Line DE falls on AB the whole Line DI shall fall on AG otherwise Two streight Lines would contain a space therefore the Angle IEF shall be equal to the Angle GBC Now if you should turn the Triangle DEF and make the Line DF to fall on AB and DE on AC because the Four Lines AB DF AC DE are equal as also the Angles A and D the Two Triangles ABC DEF shall ly exactly on each other and the Angles ACB DEF HCB IEF shall be equal Now according to our first Comparison the Angle ABC was equal to DEF and GBC to IEF therefore the Angles ABC ACB which are equal to the same DEF and GBC HCB which are also equal to the same IEF are also equal among themselves I thought fit not to make use of Euclid's Demonstration because it being too difficult for young Learners to Apprehend they are often discouraged to proceed PROPOSITION VI. THEOREM IF Two Angles of a Triangle are equal that Triangle shall be an Isosceles Triangle Let the Angles ABC ACB of the Triangle ABC be equal I affirm that their opposite Sides AB AC are equal to each other let the Triangle DEF have the Base DF equal to BC and the Angle DEF equal to ABC as also DEF equal to ACB since then that we suppose that the Angles ABC ACB are equal the Four Angles ABC ACB DEF DFE are equal Now let us imagine the Base EF be so put on BC that the Point E fall on B it is evident that since they be equal the one shall not exceed the other anywise moreover the Angle E being equal to the Angle B and the Angle F to the Angle C the Line EB shall fall on BA and FD on CA therefore the Lines ED FB shall meet each other in the Point A from whence I conclude that the Line ED is equal to BA Again let us turn over the Triangle DEF placing the Point E on C and F on B which must so happen because BC is supposed equal to EF and because the Angles F and B E and C are supposed equal the Side FB shall fall on BA and ED on CA and the Point D on A wherefore the Lines AC DE shall be equal from whence I conclude that the Sides AB AC are equal to each other since they are equal to the same Side DE. USE Use 6. THis Proposition is made use of in measuring all inaccessible Lines it is said that Thales was the first that measured the Height of the Egyptian Pyramids by their Shadows it may easily be done by this Proposition for if you would measure the Height of the Pyramid AB you must wait till the Sun be elevated 45 degrees above the Horizon that is to say untill the Angle AGB be 45 degrees now by the Proposition the shadow BC is equal to the Pyramid AB for seeing the Angle ABC is a Right Angle and that the Angle ACB is half Right or 45 degrees the Angle CAB shall also be half a Right Angle as shall be proved hereafter Therefore the Angles BCA BAC are equal and by the Sixth the Sides AB BC are also equal I might measure the same without making use of the Shadow by going backwards from B untill the Angle ACB be half a Right Angle which I may know by a Quadrant Those Propositions are made use of in Trigonometry and several other Treatises I Omit the Seventh whose use is only to Demonstrate the Eighth PROPOSITION VIII THEOREM IF Two Triangles be equally Sided they shall also have their opposite Angles equal Let the Sides GI LT HI VT GH LV be equal I say that the Angle GIH shall be equal to the Angle LTV IGH to the Angle L IHG to the Angle V. From the Center H at the distance HI let the Circle IG be described and from the Center G with the Extent GI let the Circle HI be described Demonstration If the Base LV be laid on the Base HG they will agree because they are equal I further add that the Point T shall fall precisely at the Circumference of the Circle IG since we suppose the Lines HL VT are equal it must likewise fall at the Circumference IH seeing GI is equal to LT therefore it shall fall on the Point I which is the Point were those Circles cut each other but if you deny it and say it should fall on some other Point as at O then I say the Line HO that is to say VT would be greater then HI and the Line GO that is to say LT would be less than GI which is contrary to the Hypo. from whence I conclude the Triangles do agree in all their parts and that therefore the Angles GIH is equal to the Angle LTV This Proposition is necessary for the establishing the next following moreover when we cannot measure an Angle which is made in a Solid Body by reason we cannot place our Instruments we take the Three Sides of a Triangle and make another on Paper whose Angles we easily measure this is very useful in Dialling and in cutting of Stone and Bevelling of Timber PROPOSITION IX PROBLEM TO Bisect or Divide into two equal parts a Right-Lined Angle given SAT. Take AS equal to AT and draw the Line ST upon which make an equilateral Triangle SVT draw the Right Line VA it shall Bisect the Angle Demonstration AS is equal to AT and the Side AV is common and the Base SV equal to VT therefore the Angle SAV is equal to TAV which was to be done USE THis Proposition is very useful to Divide a Quadrant into Degrees it being the same thing to Divide an Angle or to Divide an Arch into two equal parts for the Line AV Divideth as well the Arch ST at the Angle SAT for if you put the Semi-Diameter on a Quadrant you cut off an Arch of 60 Degrees and Dividing that Arch into Two Equal parts you have 30 Degrees which being again Divided into two equal parts you have 15 Degrees It is true to make an end of this Division you must Divide an Arch in Three which is not yet known to Geometricians Our Sea Compasses are Divided into 32 Points by this Proposition PROPOSITION X. PROBLEM TO Divide a Line given into two equal parts Let the Line AB be proposed to be Divided into Two equal parts erect an Equilateral Triangle ABC
BED DEC by the 4th I might prove by the same manner of Argument that BF CF are equal Demonstration The Angles BED DEC are equal it is also supposed that the Angles BED AEF are equal Thence the opposite Angles DEC AEF shall be equal and by the Cor. of the 15th Prop. AEC is a streight Line By consequence AFC is a Triangle in which the Sides AF FC are greater than AEC that is to say AE EB For the Lines AE FC are equal to AF EB Therefore the Lines AF FB are greater then AE BE and since natural Causes do act by the most shortest Lines therefore all reflections are made after this sort that the Angles of Reflection and of Incidence are equal Prop. XV. Moreover because we can easily prove that all the Angles which are made upon a Plain by the meeting of never so many Lines in the same Point are equal to Four Right Angles since that in the first Figure of this Proposition the Angles AEC AED are equal to two Right as also BED BEC we make this General Rule to determine the number of Polygons which may be joyned together to Pave a a Floor so we say that four squares six Triangles three Hexigones may Pave the same and that it is for this reason the Bees make their cellules Hexagonal PROPOSITION XVI THEOREM THe exteriour Angle of a Triangle is greater then either of the inward and Opposite Angles Continue the Side BC of the Triangle ABC I say that the exteriour Angle ACD is greater then the interiour opposite Angle ABC or BAC Imagine that the Triangle ABC moveth along BD and that it is Transported to CED Demonstration It is impossible that the Triangle ABC should be thus moved and that the Point A should not alter its place moving towards E Now if it moveth towards E the Angle ECD that is to say ABC is less than the Angle ACD therefore the Interiour Angle ABC is less than the Exteriour ACD It is easy to prove that the Angle A is also less than the Exteriour ACD For having continued the Side AC to F the opposite Angles BCF ACD are equal by the 15th and making the Triangle ABC to slide along the Line ACF it is demonstrable that the Angle BCF is greater than the Angle A. USE Use 16. WE draw from this Proposition several very useful conclusions The First that from a given Point there cannot be drawn no more than one Perpendicular to the same Line Example Let the Line AB be Perpendicular to the Line BC I say AC shall not be Perpendicular thereto because the Right Angle ABC shall be greater than the Interiour ACB Therefore ACB shall not be a Right Angle neither AC a Perpendicular Secondly that from the same Point A there can only be drawn two equal Lines for Example AC AD and that if you draw a Third AE it shall not be equal For since AC AD are equal the Angles ACD ADC are equal by the 5th now in the Triangle AEC the Exteriour Angle ACB is greater than the Interiour AEC and thus is the Angle ADE greater than AED Therefore the Line AE greater than AD and by consequence AC AE are not equal Thirdly if the Line AC maketh the Angle ACB acute and ACF obtuse the Perpendicular drawn from A shall fall on the same Side which the Acute Angle is of for if one should say that AE is a Perpendicular and the Angle AEF is Right then the Right Angle AEF would be greater than the Obtuse Angle ACE Those conclusions we make use of to measure all Parallelograms Triangles and Trapeziams and to Reduce them into Rectangular Figures PROPOSITION XVII THEOREM TWo Angles of any Triangle are less than Two Right Angles Let ABC be the proposed Triangle I say that Two of the Angles taken together BAC BCA are less than Two Right Angles Continue the Side CA to D. Demonstration The Interiour Angle C is less than the Exteriour BAD by the 16th Add to both the Angle BAC the Angles BAC BCA shall be less than the Angles BAC BAD which latter are equal to Two Right by the 13th Therefore the Angles BAC BCA are less than Two Right I might Demonstrate after the same manner that the Angles ABC ACB are less than two Right by continuing the Side BC. Coroll If one Angle of a Triangle be either Right or Obtuse the other Two shall be Acute This Proposition is necessary to Demonstrate those which follow PROPOSITION XVIII THEOREM THe greatest Side of every Triangle subtends the greatest Angle Let the Side BC of the Triangle ABC be greater then the Side AC I say that the Angle BAC which is opposite to the Side BC is greater than the Angle B which is opposite to the Side AC Cut off from BC the Line CD equal to AC and draw AD. Demonstration Seeing that the Sides AC CD are equal the Triangle ACB is an Isosceles Triangle by the 5th the Angles CDA CAD are therefore equal Now the whole Angle BAC is greater than the Angle CAD Thence the Angle BAC is greater than the Angle CDA which being exteriour in regard of the Triangle ABD is greater than the interiour B by the 16th Therefore the Angle BAC is greater than the Angle B. PROPOSITION XIX THEOREM IN every Triangle the greatest Angle is opposite to the greatest Side Let the Angle A of the Triangle BAC be greater than the Angle ABC I say that the Side BC opposite to the Angle A is greater then the Side AC opposite to the Angle B. Demonstration If the Side BC was equal to the Side AC in this case the Angles A and B would be equal by the 5th which is contrary to the Hypothesis If the Side BC was less than AC then the Angle B would be greater than A which is also contrary to the Hypothesis Wherefore I conclude that the Side BC is greater than AC USE Use 19. WE prove by these Propositions not only that from the same Point to a Line given there can be but one Perpendicular drawn but also that that Perpendicular is the shortest Line of all those Lines which might be drawn to the said Line As for instance if the Line RV be Perpendicular to ST it shall be shorter then RS because the Angle RVS being Right the Angle RSV shall be Acute by the Cor. of the 17th and the Line RV shall be shorter than RS by the Preceding For this Reason Geometricians always make use of the Perpendiculars in their measuring and Reducing irregular Figures into those whose Angles are Right I further add that seeing there can only be drawn Three Perpendiculars to one and the same Point it cannot be imagined that there are more than Three Species of quantity viz. A Line a Surface and a Solid We also prove by these Propositions that a Boul which is exactly round being put on a Plain cannot stand but on one determinate Point As for Example
Let the Line AB Represent the Plain and from the Center of the Earth C let the Line CA be drawn Perpendicular to the Line AB I say that a Boul being placed on the Point B ought not to stand on that Point because no heavy Body will stand still while it may Descend Now the Boul B going towards A is always Descending and cometh nearer and nearer the Center of the Earth C Because in the Triangle CAB the Perpendicular CA is shorter than BC. We also prove in like manner that a liquid Body ought to Descend from B to A and that the surface ought to be â—ound PROPOSITION XX. THEOREM THe Two Sides of a Triangle taken together are greater then the Third I say that the Two Sides TL LV of the Triangle TLV are greater than the Side TV. Some prove this Proposition by the Definition of a streight Line which is the shortest which can be drawn between Two Points Therefore the Line TV is less than the Two Lines TL LV. But we may Demonstrate the same another way continue the Side LV to R and let LR LT be equal then draw the Line RT Demonstration The Sides LT LR of the Triangle LTR are equal Therefore the Angles R and RTL equal by the 5â—h but the Angle RTV is greater than the Angle RTL Therefore the Angle RTV is greater than the Angle R and by the 17th in the Triangle RTV the Side RV that is to say the sum of the Sides LT LV are greater than the Side TV. PROPOSITION XXI THEOREM IF on the same Base you draw a lesser Triangle in a greater the Sides of the lesser shall be shorter than the greater but contain a greater Angle Let the less Triangle ADB be drawn within the greater ACB on the same Base AB I say in the first place that the Sides AC BC are greater than the Sides AD BD. Continue the Side AD unto E. Demonstration In the Triangle ACE the Sides AC CE taken together are greater than the Side AE by the 20th Add thereto the Side EB the Sides AC CEB shall be greater than the Sides AE EB Likewise in the Triangle DBE the Two Sides BE ED taken together are greater than BD and adding thereto AD the Sides AE EB shall be greater than AD BD. Moreover I say that the Angle ADB is greater than the Angle ACB For the Angle ADB is exteriour in respect of the Triangle DEB it is therefore greater than the Interiour DEB by the 16th Likewise the Angle DEB being Exteriour in respect of the Triangle ACE is greater than the Angle ACE Therefore the Angle ADB is greater than the Angle ACB USE Prop. XXI WE Demonstrate in Opticks by this Proposition that if from the Point C one should see the Base AB it would seem less than if one should see the same from the Point D according to this Principle that quantities seen under a greater Angle appear greater for which reason Vitruvius would that the Tops of very high Pillars should be made but little tapering because that their Tops being at a good distance from the Eyes will of themselves appear very much diminished PROPOSITION XXII PROBLEM TO make a Triangle having its Sides equal to Three Right Lines given provided that Two of them be greater than the Third Let it be proposed to make a Triangle having its Three Sides equal to the Three given Lines AB D E take with your Compasses the Line D and putting one Foot thereof in the Point B make an Arch Then take in your Compasses the Line E and putting one Foot in the Point A cross with the other Foot the former Arch in C Draw the Lines AC BC. I say that the Triangle ABC is what you desire Demonstration The Side AC is equal to the Line E since it is the Radius of an Arch drawn on the Center A equal in Length to the Line E likewise the Side BC is equal to the Line D Therefore the Three Sides AC BC AD are equal to the Lines E D AB USE WE make use of this Proposition to make a Figure equal or like unto another for having Divided the Figure into Triangles and making other Triangles having their Sides equal to the sides of the Triangles in the proposed figure we shall have a Figure like unto the same in all Respects But if we desire it should be only like thereunto but lesser for Example if we would have the Form of any Plain or Piece of Ground on Paper having Divided the same into Triangles and measured all their Sides we make Triangles like unto those of the Plain by the help of a Scale of equal parts from which we take the number of Parts which their Sides contain whether Feet Rods or any other measure and applying them as is here Taught PROPOSITION XXIII PROBLEM TO make an Angle equal to an Angle given in any Point of a Line Let it be proposed to make an Angle equal to EDF at the Point A of the Line AB at the Points A and D as Centers draw two Arches BC EF with the same extent of the Compasses then take the Distance EF between your Compasses put one Foot in B and cut off BC and draw AC I say that the Angles BAC EDF are equal Demonstration The Triangles ABC DEF have the Sides AB AC equal to the Sides DE DF since that the Arches BC EF were described with the same extent of the Compass they have also their Bases BC EF equal Therefore the Angles BAC DEF are equal by the 8th USE THis Problem is so necessary in Surveying Fortifications Prospective Dialling and in all other parts of the Mathematicks so that the greatest part of their Practice would be impossible if one Angle could not be made equal to another or of any number of Degrees required PROPOSITION XXIV THEOREM IF Two Triangles which have Two Sides of the one equal to the Two Sides of the other that which hath the greatest Angle hath the greatest Base Let the Sides AB DE AC DF of the Triangles ABC DEF be equal and let the Angle BAC be greater than the Angle EDF I say that the Base BC is greater than the Base EF Make the Angle EDG equal to the Angle BAC by the 23d. and the Line DG equal to AC then draw EG In the first place the Triangles ABC DEG having the Sides AB DE AC DG equal and the Angle EDG equal to BAC their Bases BC EG are equal by the 4th and the Lines DG DF being equal to AC they shall be equal amongst themselves Demonstration In the Triangle DGF the Sides DG DF being equal the Angles DGF DFG are equal by the 5th but the Angle EGF is less than DGF and the Angle EFG is greater than DFG Therefore in the Triangle EFG the Angle EFG shall be greater than the Angle EGF thence by the 18th the Line EG opposite to the greatest Angle EFG shall be
greater than EF thence BC equal to EG is greater than the Base EF. PROPOSITION XXV THEOREM OF Two Triangles which have Two Sides of the one equal to Two Sides of the other that Triangle which hath the greatest Base hath also the greatest Angle Let the Sides AB DE AC DF of the Triangles ABC DEF be equal and let the Base BC be greater than the Base EF. I say that the Angle A shall be greater than the Angle D. Demonstration If the Angle A were not greater than the Angle D it would be equal or less if equal in this case the Bases BC would be equal by the 4th If less then the Base EF would be greater than the Base BC by the 24th both which is contrary to our Hyp. These Propositions are of use to Demonstrate those which follow PROPOSITION XXVI THEOREM A Triangle which hath One Side and Two Angles equal to those of an other shall be equal thereto in every respect Let the Angles ABC DEF ACD DFE of the Triangles ABC DEF be equal and let the Sides BC EF which are between those Angles be also equal to each other I say that their other Sides are equal For Example AC DF but let it be imagined that the Side DF is greater than AC and that GF is equal to AC and draw the Line GE. Demonstration The Triangles ABC GEF have the Sides EF BC AC GF equal the Angle C is also supposed equal to the Angle F thence by the 4th the Triangles ABC GEF are equal in every respect and the Angles GEF ABC are equal but according to our first Hyp. the Angles ABC DEF were equal by this Argument the Angles DEF GEF would be equal that is to say the whole equal to a part which is impossible therefore DF cannot be greater than AC nor AC greater than DF because the same Demonstration might be made in the Triangle ABC Secondly Let us suppose that the Angles A and D C and F are equal and that the Sides BC EF which are opposite to the equal Angle A and D are also equal to each other I say the other Sides are equal for if DF be greater than AC make GF equal to AC and draw the Line GE. Demonstration The Triangles ABC GEF have the Sides EF BC FG CA equal they are then equal in every respect by the 4th and the Angles EGF BAC shall be equal but according to our Hyp. A and D were equal thus the Angles D and EGF should be equal which is impossible since that the Angle EGF being exteriour in respect of the Triangle EGD it ought to be greater than the Interiour Angle D by the 16th Therefore the Side DF is not greater than AC USE Use 26. THales made use of this Proposition to measure inaccessible Distances the Distance AD being proposed from the Point A he draws the Line AC Perpendicular to AD then placing a Semicircle at the Point C he measureth the Angle ACD then he takes an Angle equal thereto on the other side drawing the Line CB untill it meets the Line DA continued to the Point B. He Demonstrates that the Lines AD AB were equal so measuring actually the accessible Line he might know by that means the other For the Two Triangles ADC ABC have the Right Angles CAD CAB equal both the Angles ACD ACB were taken equal to each other and the Side AC is common to both Triangles Therefore by the 26th the Sides AD AB are equal LEMMA Lem. 26. A Line which is Perpendicular to one Parallel is also Perpendicular to the other Let the Lines AB CD be Parallel to each other and let EF be Perpendicular to CD I say that it is Perpendicular to AB Cut off two equal Lines CF FD at the Points C and D erect two Perpendiculars to the Line CD which shall also be equal to FE by the Definition of Parallels and draw the Lines EC ED. Demonstration The Triangles CEF FED have the Side FE common the Sides FD FC are equal the Angles at F are Right and by consequence equal Therefore by the 4th the Bases EC ED the Angles FED FEC FDE FCE are equal and those two last being taken away from the Right Angles ACE BDF leaveth the equal Angles EDB ECA Now the Triangles CAE DBE shall by the 4th have the Angles DEB CEA equal which Angles being added to the equal Angles CEF FED maketh equal Angles FEB FEA Therefore EF is Perpendicular to AB PROPOSITION XXVII THEOREM IF a Right Line falling upon Two Right Lines make the Alternate Angles equal the one to the other then are the Right Lines Parallel Let the Line EH fall on the Right Lines AB CD making therewith the Alternate Angles AFG FGD equal I say in the first place that the Lines AB CD shall not meet although continued as far as one lists For supposing they should meet in I and that FBI CDI are two streight Lines Demonstration If FBI GDI be two streight Lines then FIG is a Triangle then by the 16th the exteriour Angle AFG shall be greater than the interiour FGI Wherefore that the equality of the Angles may subsist the Lines AB CD must never meet each other But because we have Examples of some crooked Lines that never intersect which notwithstanding are not Parallels but approach continually To prove the foregoing I make another Demonstration as followeth First I say that if the Line EH falleth on the Lines AB CD maketh the Alternate Angles AFG FGD equal the Lines AB CD are Parallels that is in every part equidistant from each other for which reason the Perpendiculars shall be equal to each other Draw from G to the Line AB the Perpendicular GA and CD being taken equal to AF draw FD. Demonstration The Triangles AGF FGD have the Side GF common the Side GD was taken equal to AF it is supposed that the Angles AFG FGD are equal Therefore by the 4th AC FD are equal and the Angle GDF is equal to the Right Angle GAF Thence FD is Perpendicular Furthermore that AB is Parallel to CD for the Parallel to CD is to be drawn from the Point F and must pass through the Point A according to our Definition of Parallels which is that the Perpendiculars AG FD are equal PROPOSITION XXVIII THEOREM IF a Right Line falling upon Two Right Lines make the Exteriour Angle equal to the Interiour opposite Angle of the other on the same Side or the Two Interiour on the same Side equal to Two Right then are the Right Lines Parallel Having drawn a Figure like unto the former let the Line EH fall on AB CD make in the first place the Exteriour Angle EFB equal to the Interiour opposite Angle of the other on the same Side FGD I say that the Lines AB CD are Parallel Demonstration The Angle EFB is equal to the Angle AFG by the 15th and it being supposed that EFB is also equal to
the 1st because the Sides BC BD are equal the Angle ABC shall be double of each The Second case is when an Angle encloseth the other and the Lines making the same Angles not meeting each other as you see in the second figure the Angle BID is in the Center and the Angle BAD is at the Circumference Draw the Line AIC through the Center Demonstration The Angle BIC is double to the Angle BAC and CID double to the Angle CAD by the preceding case Therefore the Angle BID shall be double to the Angle BAD USE THere is given ordinarily a practical way to describe a Horizontal Dial by a single opening of the Compass which is grounded in part on this Proposition Secondly when we would determine the Apogaeon of the Sun and the excentricity of his Circle by Three observations we suppose that the Angle at the Center is double to the Angle at the Circumference Ptolomy makes often use of this Proposition to determine as well the excentricity of the Sun as the Moon 's epicycle The first Proposition of the Third Book of Trigonometry is grounded on this PROPOSITION XXI THEOREM THe Angles which are in the same Segment of a Circle or that have the same Arch for Base are equal If the Angles BAC BDC are in the same Segment of a Circle greater than a Semicircle they shall be equal Draw the Lines BI CI. Demonstration The Angles A and D are each of them half of the Angle BIC by the preceding Proposition therefore they are equal They have also the same Arch BC for Base Secondly let the Angles A and D be in a Segment BAC less than a semi-circle they shall notwithstanding be equal Demonstration The Angles of the Triangle ABE are equal to all the Angles of the Triangle DEC The Angles ECD ABE are equal by the preceding case since they are in the same Segment ABCD greater than a Semi-circle the Angles in E are likewise equal by the 15th of the 1st therefore the Angles A and D shall be equal which Angles have also the same Arch BFC for Base USE Prop. XXI IT is proved in Opticks that the Line BC shall appear equal being seen from A and D since it always appeareth under equal Angles We make use of this Proposition to describe a great Circle without having its Center For Example when we would give a Spherical figure to Brass Cauldrons to the end we may work thereon and to pollish Prospective or Telescope Glasses For having made in Iron an Angle BAC equal to that which the Segment ABC contains and having put in the Points B and C two small pins of Iron if the Triangle BAC be made to move after such a manner that the Side AB may always touch the Pin B and the Side AC the Pin C the Point A shall be always in the Circumference of the Circle ABCD. This way of describing a Circle may also serve to make large Astrolabes PROPOSITION XXII THEOREM QVadrilateral figures described in a Circle have their opposite Angles equal to Two Right Let a Quadrilateral or four sided figure be described in a Circle in such manner that all its Angles touch the Circumference of the Circle ABCD I say that its opposite Angles BAD BCD are equal to two Right Draw the Diagonals AC BD. Demonstration All the Angles of the Triangle BAD are equal to Two Right In the place of its Angle ABD put the Angle ACD which is equal thereto by the 21st as being in the same Segment ABCD and in the place of its Angle ADB put the Angle ACB which is in the same Segment of the Circle BCDA So then the Angle BAD and the Angles ACD ACB that is to say the whole Angle BCD is equal to Two Right USE PTolomy maketh use of this Proposition to make the Tables of Chords or Subtendents I have also made use thereof in Trigonometry in the Third Book to prove that the sides of an Obtuse Angled Triangle hath the same reason amongst themselves as the Sines of their opposite Angles PROPOSITION XXIII THEOREM TWo like Segments of a Circle described on the same Line are equal I call like Segments of a Circle those which contain equal Angles and I say that if they be described on the same Line AB they shall fall one on the other and shall not surpass each other in any part For if they did surpass each other as doth the Segment ACB the Segment ADB they would not be like And to demonstrate it draw the Lines ADC DB and BC. Demonstration The Angle ADB is exteriour in respect of the Triangle BDC Thence by the 21th of the 1st it is greater than the Angle ACB and by consequence the Segments ADB ACB containeth unequal Angles which I call unlike PROPOSITION XXIV THEOREM TWo like Segments of Circles described on equal Lines are equal If the Segments of Circles AEB CFD are like and if the Lines AB CD are equal they shall be equal Demonstration Let it be imagined that the Line CD be placed on AB they shall not surpass each other seeing they are supposed equal and then the Segments AEB CED shall be described on the same Line and they shall then be equal by the preceding Proposition USE Use 24. CVrved Lined Figures are often reduced to Right Lined by this Proposition As if one should describe Two like Segments of Circles AEC ADB on the equal sides AB AC of the Triangle ABC It is evident that Transposing the Segment AEC on ADB the Triangle ABC is equal to the figure ADBCEA PROPOSITION XXV PROBLEM TO compleat a Circle whereof we have but a part There is given the Arch ABC and we would compleat the Circle There needeth but to find its Center Draw the Lines AB BC and having Divided them in the middle in D and E draw their Perpendiculars DI EI which shall meet each other in the Point I the Center of the Circle Demonstration The Center is in the Line DI by the Coroll of the 1st It is also in EI it is then in the Point I. USE Use 25. THis Proposition cometh very frequently in use it might be propounded another way as to inscribe a Triangle in a Circle or to make a Circle pass through three given points provided they be not placed in a streight Line Let be proposed the Points A B C put the Point of the Compass in C and at what opening soever describe Two Arks F and E. Transport the foot of the Compass to B and with the same opening describe Two Arcks to cut the former in E and F. Describe on B as Center at what opening soever the Arches H and G and at the same opening of the Compass describe on the Center A Two Arks to cut the same in G and H. Draw the Lines FE and GH which will cut each other in the Point D the Center of the Circle The Demonstration is evident enough for if you had drawn
ABC Draw the touch Line FED by the 17th of the 3d. and make at the Point of touching E the Angle DEH equal to the Angle B and the Angle FEG equal to the Angle C by the 23d of the 1st Draw the Line GH the Triangle EGH shall be equi Angled to ABC Demonstration The Angle DEH is equal to the Angle EGH of the Alternate Segment by the 32d of the 3d. now the Angle DEH was made equal to the Angle B and by consequence the Angles B and G are equal The Angles C and H are also equal for the same reason and by the Coroll 2. of the 32d of the 1st the Angles A and GEH shall be equal Therefore the Triangles EGH ABC are equiangled PROPOSITION III. PROBLEM TO describe about a Circle a Triangle equiangled to another If one would describe about a Circle GKH a Triangle equiangled to ABC one of the Sides BC must be continued to D and E and make the Angle GIH equal to the Angle ABD and HIK equal to the Angle ACE then draw the Tangents LGM LKN NHM through the Points G K H. The Tangents shall meet each other for the Angles IKL IGL being Right if one should draw the Line KG which is not drawn the Angles KGL GKL would be less than two Right therefore by the 11th Axiom the Lines GL KL ought to concur Demonstration All the Angles of the Quadrilateral GIHM are equal to four Right seeing it may be reduced into Two Triangles the Angles IGM IHM which are made by the Tangents are Right thence the Angles M and I are equivalent to Two Right as well as the Angles ABC ABD Now the Angle GIH is equal to the Angle ABD by construction therefore the Angle M shall be equal to the Angle ABC For the same reason the Angles N and ACB are equal and so the Triangles LMN ABC are equiangled PROPOSITION IV. PROBLEM To inscribe a Circle in a Triangle IF you would inscribe a Circle in a Triangle ABC divide into Two equally the Angles ABC ACB by the 9th of the 1st drawing the Lines CD BD which concurr in the Point D. Then draw from the Point D the Perpendiculars DE DF DG which shall be equal so that the Circle described on the point D at the opening DE shall pass through F and G. Demonstration The Triangles DEB DBF have the Angles DEB DFB equal seeing they are Right the Angles DBE DBF are also equal the Angle ABC having been divided into Two equally the Side DB is common therefore by the 26th of the 1st these Triangles shall be equal in every respect and the Sides DE DF shall be equal One might demonstrate after the same manner that the Sides DF DG are equal One may therefore describe a Circle which shall pass through the Points E F G and seeing the Angles E F G are Right the Sides AB AC BC touch the Circle which shall by consequence be inscribed in the Triangle PROPOSITION V. PROBLEM To describe a Circle about a Triangle IF you would describe a Circle about a Triangle ABC divide the Sides AB BC into Two equally in D and E drawing the Perpendiculars DF EF which concurr in the Point F. If you describe a Circle on the Center F at the opening FB it shall pass through A and C that is to say that the Lines FA FB FC are equal Demonstration The Triangles ADF BDF have the Side DF common and the Sides AD DB equal seeing the Side AB hath been divided equally and the Angles in D are equal being Right Thence by the 4th of the 1st the Bases AF BF are equal and for the same reason the Bases BF CF. USE WE have often need to inscribe a Triangle in a Circle as in the first Proposition of the Third Book of Trigonometry This practice is necessary for to measure the Area of a Triangle and upon several other occasions PROPOSITION VI. PROBLEM To inscribe a Square in a Circle TO inscribe a Square in the Circle ABCD draw to the Diameter AB the Perpendicular DC which may pass through the Center E. Draw also the Lines AC CB BD AD and you will have inscribed in the Circle the Square ACBD Demonstration The Triangles AEC CEB have their Sides equal and the Angles AEC CEB equal seeing they are Right therefore the Bases AC CB are equal by the 4th of the 1st Moreover seeing the Sides AE CE are equal and the Angle E being Right they shall each of them be semi-right by the 32d of the 1st So then the Angle ECB is semi-right And by consequence the Angle ACB shall be Right It is the same of all the other Angles therefore the Figure ACDB is a Square PROPOSITION VII PROBLEM To describe a Square about a Circle HAving drawn the Two Diameters AB CD which cut each other perpendicularly in the Center E draw the touch Lines FG GH HI FI through the Points A D B C and you will have described a Square FGHI about the Circle ACBD Demonstration The Angle E and A are Right thence by the 29th of the 1st the Lines FG CD are parallels I prove after the same manner that CD HI FI AB AB GH are parallels thence the Figure FCDG is a parallelogram and by the 34th of the 1st the Lines FG CD are equal as also CD IH FI AB AB GH and by consequence the Sides of the Figure FG GH HI IF are equal Moreover seeing the Lines FG CD are parallels and that the Angle FCE is Right the Angle G shall be also Right by the 29th of the 1st I demonstrate after the same manner that the Angles F H and I are Right Therefore the Figure FGHI is a Square and its Sides touch the Circle PROPOSITION VIII PROBLEM To inscribe a Circle in a Square IF you will inscribe a Circle in the Square FGHI divide the Sides FG GH HI FI in the middle in A D B C and draw the Lines AB CD which cutteth each other in the Point E. I demonstrate that the Lines EA ED EC EB are equal and that the Angles in A B C D are Right and that so you may describe a Circle on the Center E which shall pass through A D B C and which toucheth the Sides of the Square Demonstration Seeing the Lines AB GH conjoyns the Lines AG BH which are parallel and equal they shall be also parallel and equal therefore the Figure AGHB is a parallelogram and the Lines AE GD AG ED being parallel and AG GD being equal AE ED shall be also equal It is the same with the others AE EC EB Moreover AG ED being parallels and the Angle G being Right the Angle D shall be also a Right Angle One may then on the Center E describe the Circle ADBC which shall pass through the Points A D B C and which shall touch the Sides of the Square PROPOSITION IX PROBLEM To describe a Circle about a Square
Triangle ABC the Line DE is Parallel to the Base BC the Sides AB AC shall be divided proportionally that is to say that there shall be the same Reason of AD to DB as of AE to EC Draw the Lines DE BE. The Triangles DBE DEC which have the same Base DE and are between the same Parallels DE BC are equal by the 37th of the 1st Demonstration The Triangles ADE DBE have the same point E for their vertical if we take AD DB for their Bases and if one should draw through the point E a Parallel to AB they would be both between the same Parallels they shall have thence the same Reason as their Bases by the 1st that is to say that there is the same Reason of AD to DB as of the Triangle ADE to the Triangle DBE or to its equal CED Now there is the same Reason of the Triangle ADE to the Triangle CED as of the Base AE to EC There is therefore the same Reason of AD to DB as of AE to EC And if there be the same Reason of AE to EC as of AD to DB I say that the Lines DE BC would then be Parallels Demonstration There is the same Reason of AD to DB as of the Triangle ADE to the Triangle DBE by the 1st there is also the same Reason of AE to EC as of the Triangle ADE to the Triangle DEC consequently there is the same Reason of the Triangle ADE to the Triangle BDE as of the same Triangle ADE to the Triangle CED So then by the 7th of the 5th the Triangles BDE CED are equal And by the 39th of the 1st they are between the same Parallels USE THis Proposition is absolutely necessary in the following Propositions one may make use thereof in Measuring as in the following figure If it were required to measure the height BE having the length of the staff DA there is the same Reason of CD to DA as of BC to BE. PROPOSITION III. THEOREM THat Line which divideth the Angle of a Triangle into two equal parts divideth its Base in two parts which are in the same Reason to each other as are their Sides And if that Line divideth the Base into parts proportional to the Sides it shall divide the Angle into Two equally If the Line AD divideth the Angle BAC into Two equal parts there shall be the same Reason of AB to AC as of BD to DC Continue the Side CA and make AE equal to AB then draw the Line EB Demonstration The exterior Angle CAB is equal to the Two interior Angles AEB ABE which being equal by the 5th of the 1st seeing the Sides AE AB are equal the Angle BAD the half of BAC shall be equal to one of them that is to say to the Angle ABE Thence by the 27th of the 1st the Lines AD EB are parallel and by the 2d there is the same Reason of EA or AB to AC as of BD to DC Secondly If there be the same Reason of AB to AC as of BD to DC the Angle BAC shall be divided into Two equally Demonstra There is the same reason of AB or AE to AC as of BD to DC thence the Lines EB AD are parallel and by the 29th of the 1st the Alternate Angles EBA BAD the internal BEA and the external DAC shall be equal and the Angles EBA AEB being equal the Angles BAD DAC shall be so likewise Wherefore the Angle BAC hath been divided equally USE WE make use of this Proposition to attain to the Proportion of the sides PROPOSITION IV. THEOREM EQuiangular Triangles have their Sides Proportional If the Triangles ABC DCE are equiangular that is to say that the Angles ABC DCE BAC CDE be equal There will be the same Reason of BA to BC as of CD to CE. In like manner the reason of BA to AC shall be the same with that of CD to DE. Joyn the Triangles after such a manner that their Bases BC CE be on the same Line and continue the sides ED BA seeing the Angles ACB DEC are equal the Lines AC EF are parallel and so CD BF by the 29th of the 1st and AF DC shall be a parallelogram Demonstration In the Triangle BFE AC is parallel to the Base FE thence by the 2d there shall be the same reason of BA to AF or CD as of BC to CE and by exchange there shall be the same reason of AB to BC as of DC to CE. In like manner in the same Triangle CD being parallel to the Base BF there shall be the same Reason of FD or AC to DE as of BC to GE by the 2d and by exchange there shall be the same reason of AC to BC as of DE to CE. USE THis Proposition is of a great extent and may pass for a universal Principle in all sorts of Measuring For in the first place the ordinary practice in measuring inaccessible Lines by making a little Triangle like unto that which is made or imagined to be made on the ground is founded on this Proposition as also the greatest part of those Instruments on which are made Triangles like unto those that we would measure as the Geometrical Square Sinical Quadrant Jacobs Staff and others Moreover we could not take the plane of a place but by this Proposition wherefore to explain its uses we should be forced to bring in the first Book of practical Geometry PROPOSITION V. THEOREM TRiangles whose sides are proportional are equianguler If the Triangles ABC DEF have their sides proportional that is to say if there be the same reason of AB to BC as of DE to EF as also if there be the same reason of AB to AC as of DE to DF the Angles ABC DEF A and D C and F shall be equal Make the Angle FEG equal to the Angle B and EFG equal to the Angle C. Demonstration The Triangles ABC EFG have two Angles equal they are thence equiangled by the Cor. of the 32d of the 1st and by the 4th there is the same reason of DE to EF as of EG to EF. Now it is supposed that there is the same reason of DE to EF as of EG to EF. Thence by the 7th of the 5th DE EG are equal In like manner DF FG are also equal and by the 8th of the 1st the Triangles DEF GEF are equiangular Now the Angle GEF was made equal to the Angle B thence DEF is equal to the Angle B and the Angle DFE to the Angle C. So that the Triangles ABC DEF are equiangular PROPOSITION VI. THEOREM TRiangles which have their sides proportional which include an equal Angle are equiangular If the Angles B and E of the Triangles ABC DEF being equal there be the same reason of AB to BC as of DE to EF the Triangles ABC DEF shall be equiangular Make the Angle FEG equal to the Angle B and
the Angle EFG equal to the Angle C. Demonstra The Triangles ABC EGF are equiangular by the Cor. of the 32d of the 1st there is thence the same reason of AB to BC as of EG to EF by the 4th Now as AB to BC so is DE to EF there is then the same reason of DE to EF as of GE to EF. So then by the 7th of the 5th DE EG are equal and the Triangles DEF GEF which have their Angles DEF GEF each of them equal to the Angle B and the Sides DE EG equal with the Side EF common they shall be equal in every respect by the 4th of the 1st they are thence equiangular and the Triangle EGF being equal to the Triangle ABC the Triangles ABC DEF are equiangular The Seventh Proposition is unnecessary PROPOSITION VIII THEOREM A Perpendicular being drawn from the Right Angle of a Right Angled Triangle to the opposite side divideth the same into Two Triangles which are a like thereto If from the Right Angle ABC be drawn a perpendicular BD to the opposite side AC it will divide the Right Angled Triangle ABC into Two Triangles ADB BDC which shall be like or equiangular to the Triangle ABC Demonstration The Triangles ABC ADB have the same Angle A the Angle ADB ABC are right they are thence equiangular by the Cor. 2. of the 32d of the 1st In like manner the Triangles BDC ABC have the Angle C common and the Angles ABC BDC being right they are also equal Thence the Triangles ABC DBC are like USE WE measure inaccessable distances by a Square according to this Proposition For example if we would measure the distance DC having drawn the perpendicular DB and having put a Square at the Point B in such manner that by looking over one of its Sides BC I see the Point C and over its other Side I see the Point A it is evident that there will be the same reason of AD to DB as of DB to DC So that multiplying DB by its self and dividing that product by AD the Quotient shall be DC PROPOSITION IX PROBLEM To cut off from a Line any part required LEt there be proposed the Line AB from which it is required to cut off three Fifths Make the Angle ECD at discretion take in one of those Lines CD five equal parts and let CF be three of the same and CE be equal to AB Then draw the Line DE after which draw FG parallel to DE the Line CG will contain three Fifths of CE or AB Demonstration In the Triangle ECD FG being parallel to the Base DE there will be the same Reason of CF to FD as of CG to GE by the second and by composition by the 18th of the 5th there shall be the same Reason of CG to CE as of CF to CD Now CF contains three fifths of CD wherefore CG shall contain three fifths of CE or AB PROPOSITION X. PROBLEM TO divide a Line after the same manner as another Line is divided If one would divide the Line AB after the same manner as the Line AC is divided Joyn those Lines making an Angle at pleasure as CAB Draw the Line BC and the parallels EO FV and the Line AB shall be divided after the same manner as AC Demonstration Seeing that in the Triangle BAC the Line HX hath been drawn parallel to the Base BC it will divide the Sides AB AC proportionally by the second it is the same with all the other parallels To do the same with more facility one may draw BD parallel to AC and put off the same Divisions of AC on BD then draw the Lines from the one to the other PROPOSITION XI THEOREM TO find a third Proportional to Two given Lines It is required to find a third proportional to the Lines AB BC that is to say that there may be the same reason of AB to BC as of BC to the Line required Make at discretion the Angle EAC put off one after the other the Lines AB BC and let AD be equal to BC Draw the Lines BD and its parallel CE. The Line DE shall be that which you require Demonstration In the Triangle EAC the Line DB is parallel to the Base CE There is thence by the 2d the same reason of AB to BC as of AD or BC to DE. PROPOSITION XII PROBLEM TO find a fourth Proportional to three Lines given Let there be proposed three Lines AB BC DE to which must be found a fourth proportional make an Angle as FAC at discretion take on AC the Lines AB BC and on AF the Line AD equal to DE then draw DB and its parallel FC I say that DF is the Line you seek for that is to say that there is the same Reason of AB to BC as of DE or AD to DF. Demonstration In the Triangle FAC the Line DB is parallel to the Base FC there is thence the same reason of AB to BC as of AD to DF by the 2d USE THe use of the Compass of Proportion or Sector is established on these Propositions for we divide a Line as we please by the Compass of Proportion we do the Rule of Three without making use of Arithmetick we extract the Square Root and Cube Root we double the Cube we measure all sorts of Triangles we find the Content of Superficies and the solidity of Bodies we augment or diminish any figure whatever according to what Proportion we please and all those uses are Demonstrated by the foregoing Propositions PROPOSITION XIII PROBLEM TO find a mean Proportional between Two Lines If you would have a mean Proportional between the Lines LV VR having joyned them together on a strait Line divide the Line LR into two equal parts in the point M and having described a Semi-circle LTR on the Center M draw the perpendicular VT it shall be a mean Proportional between LV VR Draw the Lines LT TR. Demonstration The Angle LTR described in a Semi-circle is right by the 31st of the 3d. and by the 8th the Triangles LVT TVR are like there is thence the same Reason in the Triangle LVT of LV to VT as of VT to VR in the Triangle TVR by the 4th So then VT is a mean Proportional between LV and VR USE WE Reduce to a Square any Rectangular Parallelogram whatever by this Proposition For example in the Rectangle comprehended under LV VR I will demonstrate hereafter that the Square of VT is equal to a Rectangle comprehended under LV and VR PROPOSITION XIV THEOREM EQuiangular and equal Parallelograms have their Sides reciprocal and equiangular Parallelograms whose sides are reciprocal are equal If the Parallelograms L and M be equiangular and equal they shall have their sides reciprocal that is to say that there shall be the same Reason of CD to DE as of FD to DB. For seeing they have their Angles
equal they may be joyned after such manner that their joyned sides CD DE be on one streight Line by the 15th of the 1st continue the Sides AB GE you will have compleated the Parallelogram BDEH Demonstration Seeing the Parallelogram L and M are equal they shall have the same reason to the Parallelogram BDEH Now the reason of the parallelogram L to the Parallelogram BDEH is the same with that of the Base CD to the Base DE by the 1st and that of the Parallelogram M or DFGE is the same with that of the Base FD to the Base BD. Thence there is the same reason of CD to DE as of FD to BD. Secondly if the equiangular Parallelograms L and M have their Sides reciprocal they shall be equal Demonstration The Sides of the Parallelograms are reciprocal that is to say that there is the same Reason of CD to DE as of FD to BD Now as the Base CD is to DE so is the parallelogram L to the parallelogram BDEH by the first and as FD is to DB so is the parallelogram M to BEDH there is thence the same Reason of L to BDEH as of M to the same BDEH so then by the 7th of the 5th the parallelograms L and M are equal PROPOSITION XV. THEOREM EQual Triangles which have one Angle equal have the Sides which form that Angle reciprocal and if their sides be reciprocal they shall be equal If the Triangles F and G being equal have their Angles ACB ECD equal their sides about that Angle shall be reciprocal that is to say that there shall be the same Ratio of BC to CE as of CD to CA. Dispose the Triangles after such a manner that the Sides CD CA be one straight Line seeing the Angles ACB ECD are supposed equal the Lines BC CE shall be also a straight Line by the 14th of the 1st Draw the Line AE Demonstration There is the same Ratio of the Triangle ABC to the Triangle ACE as of the Triangle ECD equal to the first to the same Triangle ACE by the 7th of the 5th Now as ABC is to ACE so is the Base BC to the Base CE by the 1st seeing they have the same vertical A and as FCD is to ACE so is the Base CD to CA. Now if it be supposed that the sides are reciprocal that is to say that there be the same Ratio of BC to CE as of CD to CA the Triangles ABC CDE shall be equal because they would then have the same Ratio to the Triangle ACE PROPOSITION XVI THEOREM IF four Lines be proportional the Rectangle comprehended under the first and fourth is equal to the Rectangle comprehended under the second and third and if the Rectangle comprehended under the extrems be equal to the Rectangle comprehended under the means then are the four Lines proportional if the Lines AB CD be proportional that is to say that there is the same Ratio of A to B as of C to D the Rectangle comprehended under the first A and the fourth D shall be equal to the Rectangle comprended under B and C. Demonstration The Rectangles have their Angles equal seeing it is Right they have also their Sides reciprocal they are thence equal by the 16th In like manner if they be equal their sides are reciprocal that is to say that there is the same Ratio of A to B as of C to D. PROPOSITION XVII THEOREM IF three Lines be proportional the Rectangle comprehended under the first and third is equal to the Square of the mean and if the Square of the mean be equal to the Rectangle of the extrems the three Lines are then proportional If the three Lines A B D be proportional the Rectangle comprehended under A and under D shall be equal to the Square of B. Take C equal to B there shall be the same Ratio of A to B as of C to D thence the four Lines A B C D are proportional Demonstration The Rectangle under A and D shall be equal to the Rectangle under B and C by the foregoing Now this last Rectangle is a Square seeing the Lines B and C are equal thence the Rectangle under A and D is equal to the Square of B. In like manner if the Rectangle under A and D be equal to the Square of B there shall be the same Ratio of A to B as of C to D and seeing that B and C are equal there shall be the same Ratio of A to B as of B to D. USE THose Four Propositions demonstrateth that Rule of Arithmetick which we commonly call the Rule of Three and consequently the Rule of Fellowship false position and all other which are performed by Proportion For example let there be proposed these three Numbers A Eight B six C four and it is required to find the fourth proportional Suppose it to be found and let it be D. The Rectangle comprehended under A and D is equal to to the Rectangle comprehended under B and C. Now I can have this Rectangle by Multiplying B by C that is to say six by four and I shall have twenty four thence the Rectangle comprehended under A and D is twenty four wherefore dividing the same by A eight the Quotient three is the number I look for PROPOSITION XVIII THEOREM TO describe a Poligon like to another on a Line given There is proposed the Line AB on which one would describe a Poligon like unto the Poligon CFDE Having divided the Poligon CFDE into Triangles make on the Line AB a Triangle ABH like unto the Triangle CFE that is to say make the Angle ABH equal to the Angle CFE and BAH equal to FCE So then the Triangles ABH CFE shall be equiangled by the 32d of the first make also on BH a Triangle equiangled to FDE Demonstration Seeing the Triangles which are parts of the Poligons are equiangular the two Poligons are equiangular Moreover seeing the Triangles ABH CFE are equiangular there is the same Ratio of AB to BH as of CF to FE by the 4th In like manner the Triangles HBG EFD being equiangular there shall be the same Ratio of BH to BG as of FE to FD and by equality there shall be the same Ratio of AB to BG as of CF to FD. And so of the rest of the sides Thence by the first Definition the Poligons are like to each other USE IT is on this Proposition we establish the greatest part of the practical ways to take the plane of a place of an Edifice of a Field of a Forrest or of a whole Country for making use of the equal parts of a Line for Feet or for Chains we describe a figure like unto the Prototype but lesser in which we may see the Proportion of all its Lines And because it is easier on paper than on the ground we may comprehend in this Proposition all Geodesâ—… all Chorography all Geographical Charts and ways of reducing of the
Square of the Line added are double to the Square of half the Line and to the Square which is Composed of the half Line and the Line added If one supposeth AB to be Divided in the middle at the Point C and if thereto be added the Line BD the Squares of AB and BD shall be double to the Squares of AC and CD added together Draw the Perpendiculars CE DF equal to AC Then draw the Lines AE EF AG EBG Demonstration The Lines AC CE CB beng equal and the Angles at the Point C being Right The Angles AEC CEB CBE DBG DGB shall be half Right and the Lines DB DG and EF FG CD shall be equal The Square of AE is double to the Square of AC the Square of EG is double to the Square of EF or CD by the 47th of the 1st Now the Square of AG is equal to the Squares of AE EG by the 47th of the 1st Therefore the Square of AG is double to the Squares of AC CD The same AG by the 47th of the 1st is equal to the Squares of AD BD or GD Therefore the Squares of AD BD are double the Squares of AC CD ARITHMETICALLY LEt AB be 6 parts AC 3 CB 3 BD 4 the Square of AD 10 is 100. the Square of BD 4 is 16 which are 116. The Square of AC 3 is 9 the Square of CD 7 is 49. Now 49. and 9 is 58 the half of 116. PROPOSITION XI PROBLEM TO Divide a Line so that the Rectangle comprehended under the whole Line and under one of its parts shall be equal to the Square of the other part It is proposed to Divide the Line AB so that the Rectangle comprehended under the whole Line AB and under HB be equal to the Square of AH Make a Square of AB by the 46th of the 1st Divide AD in the middle in E then Draw EB and make EF equal to EB Make the Square AF that is to say that AF AH be equal I say that the Square of AH shall be equal to the Rectangle HC comprehended under HB and the Line BC equal to AB Demonstration The Line AD is equally divided in E and there is added thereto the Line FA thence by the 6th the Rectangle DG comprehended under DF and FG equal to AF with the Square of AE is equal to the Square of EF equal to EB Now the Square of EB is equal to the Squares of AB AE by the 47th of the 1st therefore the Squares of AB AE are equal to the Rectangle DG and to the Square of AE and taking away from both the Square of AE the Square of AB which is AC shall be equal to the Rectangle DG taking also away the Rectangle DH which is in both the Rectangle HC shall be equal to the Square of AG. USE THis Proposition serveth to cut a Line in extream and mean Proportion as shall be shewn in the Sixth Book It is used often in the 14th of Euclid's Elements to find the Sides of Regular Bodies It serveth for the 10th of the Fourth Book to inscribe a Pentagone in a Circle as also a Pentadecagone You shall see other uses of a Line thus divided in the 30th of the Sixth Book PROPOSITION XII THEOREM IN an obtuse angled Triangle the Square of the side opposite to the obtuse Angle is equal to the Squares of the other two sides and to two Rectangles comprehended under the side on which one draweth a Perpendicular and under the Line which is between the Triangle and that perpendicular Let the Angle ACB of the Triangle ABC be obtuse and let AD be drawn perpendicular to BC. The Square of the side AB is equal to the Squares of the sides AC CB and to two Rectangles comprehended under the side BC and under DC Demonstration The Square of AB is equal to the Squares of AD DB. by the 47th of the 1st the Square of DB is equal to the Squares of DC and CB and to two Rectangles comprehended under DC CB by the 4th therefore the Square of AB is equal to the Squares of AD DC CB and to two Rectangles comprehended under DC CB in the place of the two last Squares AD DC Put the Square of AC which is equal to them by the 47th of the 1st the Square of AB shall be equal to the Square of AC and CB. and to two Rectangles comprehended under DC CB. USE THis Proposition is useful to measure the Area of a Triangle it s three sides being known for Example If the side AB was twenty Foot AC 13 BC 11 the Square of AB would be four hundred the Square of AC one hundred sixty nine and the Square of BC one hundred twenty one the Sum of the two last is Two hundred and ninety which being subtracted from four hundred leaves one hundred and ten for the two Rectangles under BC CD the one half-fifty five shall be one of those Rectangles which divided by BC 11 we shall have five for the Line CD whose Square is twenty five which being subtracted from the Square of AC one hundred sixty nine there remains the Square of AD one hundred forty four and its Root shall be the side AD which being multiplied by 5½ the half of BC you have the Area of the Triangle ABC containing 66 square Feet PROPOSITION XIII THEOREM IN any Triangle whatever the Square of the side opposite to an acute Angle together with two Rectangles comprehended under the side on which the Perpendicular falleth and under the Line which is betwixt the Perpendicular and that Angle is equal to the Square of the other sides Let the proposed Triangle be ABC which hath the Angle C acute and if one draw AD perpendicular to BC the Square of the side AB which is opposite to the acute Angle C together with two Rectangles comprehended under BC DC shall be equal to the Squares of AC BC. Demonstration The Line BC is divided in D whence by the 7th the Square of BC DC are equal to two Rectangles under BC DC and to the Square of BD add to both the Square of AD the Square of BD DC AD shall be equal to two Rectangles under BC DC and to the Squares of BD AD in the place of the Squares of CD AD put the Square of AC which is equal to them by the 47th of the 1st and instead of the Squares of BD AB substitute the Square of AB which is equal to them the Squares of BC AC shall be equal to the Square of AB and to two Rectangles comprehended under BC DC USE THese Propositions are very necessary in Trigonometry I make use thereof in the eighth Proposition of the third Book to prove That in a Triangle there is the same Reason between the whole Sine and the Sine of an Angle as are between the Rectangle of the sides comprehending that Angle and
will then at last leave a lesser quantity than G. For it is evident that having proposed Two unequal Quantaties if you take away more than the half of the greater and again more than the half of the remainder and again more than the half of that remainder and so forward that which remains shall be less than the second quantity Let us suppose that the second is contained One Hundred times in the first it is evident that by Dividing the first into One Hundred parts in such sort that the first may have a greater Reason to the second than of Two to One the second shall be less than the Hundreth part So then you shall at length meet with a Polygon which shall be less surpassed by the Circle than the Circle doth the figure A that is to say that which remains of the Circle having taken away the Polygon shall be less than G. The Polygon shall be greater than the figure A. PROPOSITION II. THEOREM CIrcles are in the same Ratio as are the Squares of their Diameters I Demonstrate that the Circles A and B are in the same Ratio as are the Squares of CD EF. For if they were not in the same Ratio the Circle A would have a greater Ratio to the Circle B than the Square of CD to the Square of EF. Let the figure G have the same Ratio to the Circle B as hath the Square of CD to the Square of EF the figure G shall be lesser than the Circle A by the preceding Lemma there may be inscribed a Regular Polygon greater than G in the Circle A. Let there also be inscribed in the Circle B a like regular Polygon Demonstration The Polygon A to the Polygon B hath the same Ratio as the Square of CD to the Square of EF that is to say the same as hath G to the Circle B Now the quantity G is lesser than the Polygon inscribed in A so then by the 14th of the 5th the Circle should be less than the Polygon which is inscribed which is evidently false It must then be said that the figure G less than the Circle A cannot have the same Ratio to the Circle B as the Square of CD to the Square of EF and by consequence that Circle A hath not a greater Ratio to the Circle B than the Square of CD to the Square of EF. Neither hath it less because that the Circle B to the Circle A would have a greater Ratio and there would be applyed the same Demonstration Coroll 1. Circles are in duplicate Ratio of that of their Diameters because that Squares being like or similar are in duplicate Ratio of that of their sides Coroll 2. Circles are in the same Ratio as are the similar Polygons inscribed in them Coroll 3. This general Rule must be well taken notice of when like figures inscribed in other like figures in such sort that they become nearer and nearer and degenerate in fine into those figures they are always in the same Ratio I would say that if like regular Polygons be described in several Circles they are in the same Ratio as are the Squares of the Diameters and that the greater number of sides they are made to have they become so much the nearer the Circles the Circle shall have the same Ratio as the Squares of their Diameters This way of measuring Bodies by inscription is very necessary USE THis Proposition is very universal and is the way of our reasoning on Circles after the same manner as on Squares For example we say in the 47th of the first that in a Rectangular Triangle that the Square of the Base is equal to the Square of the other sides taken together We may say the same of Circles that is to say that the Circle described on the Base of a Rectangular Triangle is equal to the Circles which hath the sides for Diameters and after this we may augment or diminish a Circle into what Proportion we list We prove also in Opticks that the Light decreaseth in duplicate Ratio of that of the distance of the Luminous Body PROPOSITION III. THEOREM EVery Pyramid whose Base is Triangular may be divided into two equal Prisms which are greater than half of the Pyramid and into Two equal Pyramids There may be found in the Pyramid ABCD two equal Prisms EBFI EHKC which shall be greater than half the Pyramid Divide the six Sides of the Pyramid into equal parts in G F E I H K and draw the Lines EG GF FE HI IK EK Demonstration In the Triangle ABD there is the same Ratio of AG to GB as of AF to FD seeing they are equal thence by the 2d of the 6th GF BD are parallels and GF shall be the half of BD that is to say equal to BH In like manner GE BI FE HI shall be parallel and equal and by the 15th of the 11th the Planes GFE BHI shall be parallel and by consequence EBFI shall be a Prism I say the same of the figure HEKF which shall also be a Prism equal to the former and by the 40th of the 11th seeing the Parallelogram Base HIKD is double to the Triangle BHI by the 41th of the first Secondly the Pyramids AEFG ECKI are like and equal Demonstration The Triangles AFG FDH are equal by the 3d. of the first as also FDH EIK. In like manner the Triangles AGE EIC and so of the other Triangles of the Pyramid they are then equal by the 10th and 11th Def. they are also similar to the great Pyramid ABDC for the Triangles ABC AGE are like by the 2d of the 6th the Lines GE being parallel which I could Demonstrate in all the Triangles of the lesser Pyramids In fine I conclude that the Prisms are more than the one half of the first Pyramid For if each were equal to one of the lesser Pyramids the Two Prisms would be the half of the great Pyramid Now they are greater than one of the Pyramids as the Prism GHE contains a Pyramid AGFE which may easily be proved from the parallelism of their Sides whence I conclude that the Two Prisms taken together are greater than the Two Pyramids and consequently greater than half the greater Pyramid PROPOSITION IV. THEOREM IF Two Triangular Pyramids of equal height be divided into Two Prisms and Two Pyramids and that the last Pyramids be divided after the same manner all the Prisms of the one Pyramid shall have the same Ratio to all those of the other as the Base of the one Pyramid hath to the Base of the other If one divide the Two Pyramids ABCD DEFG of equal height and of Triangular Bases into Two Prisms and into Two Pyramids according to the method of the Third Proposition and if one should subdivide after the same manner the Two little Pyramids and so consecutively in such sort that there be as many divisions in the one as in the other there then being the same number
side 13. The streight Lines have not the same common Segment Fig. 20. I would say that of two streight Lines AB CD which meeteth each other in the Point B is not made one single Line BD but that they cut each other and separate after their so meeting For if a Circle be described on the Center B AFB shall be a Semicircle seeing the Line ABD passing through the Center B divideth the Circle into two equally the Segment CFD should be also a Semicircle if CBD were a streight Line because it passeth through the Center B therefore the Segment CFD should be equal to the Segment AFD a part as great as the whole which would be contrary to the Ninth Axiom ADVERTISEMENT WE have two sorts of Propositions some whereof considereth only a truth without descending to the practice thereof and we call those Theorms The other proposeth something to be done or made and are called Problems The first numbers of Citations is that of the Proposition the second that of the Book As by the 2 of the 3 is signified the second proposition of the third Book but if one meet with one number thereby is meant the Proposition of the Book whose Explication is in hand PROPOSITION I. PROBLEM UPon a finite Right Line AB to describe an Equilateral Triangle ACB From the Centers A and B at the distance of AB describe the Circles cutting each other in the Point C from whence draw two Right Lines CA CB thence are AC AB BC AC equal wherefore the Triangle ACB is equilateral which was to be done USE EUclid has not applyed this Proposition to any other use but to demonstrate the two following Propositions but we may apply it to the measuring of an inaccessible Line Use 1. As for Example let AB be an inaccessable Line which is so by reason of a River or some other Impediment make an Equaliteral Triangle as BDE on Wood or Brass or on some other convenient thing which having placed Horizontally at a station at B look to the Point A along the Side BD and to some other Point C along the side BE then carry your Triangle along the Line BC so far that is untill such time as you can see the Point B your first station by the side CG and the Point A by the Side CE I say that then the Lines CB and CA are equal wherefore if you measure the Line BC you will likewise know the length of the Line AB PROPOSITION II. PROBLEM AT a Point given A to make a Right Line AC equal to a Right Line given BC. From the Center C at the distance CB describe the Circle CBE joyn AC upon which raise the Equilateral Triangle ADC produce DC to E from the Center D and the distance DE describe the Circle DEH Let DA be produced to the Point G in the Circumference thereof then AG is equal to CB for DG is equal to DE and DA to DC wherefore AG CE BC AG are equal which was to be done SCHOL THe Line AG might be taken with a pair of Compasses but the so doing answers no Postulate as Proclus well intimates PROPOSITION III. PROBLEM TWo Right Lines A and BC being given from the greater BC to take away the Right Line BE equal to the lesser A. At the Point B draw the Right Line BD equal to A the Circle described from the Center B at the distance BD shall cut off BE equal to BD equal to A equal to BE which was to be done The use of the two preceding Propositions are very evident since we are obliged very often in our Geometrical Practices to draw a Line equal to a Line given or to take away from a greater Line given a part equal to a lesser PROPOSITION IV. THEOREM IF Two Triangles ABC EDF have two Sides of the one BA AC equal to two Sides of the other ED DF each to his correspondent side that is BA to ED and AC to DF and have the Angle A equal to the Angle D contained under the equal Right Lines they shall have the Base BC equal to the Base EF and the Triangle BAC shall be equal to the Triangle EDF and the remaining Angles B C shall be equal to the remaining Angles E F each to each which are subtended by the equal Sides If the Point D be applied to the Point A and the Right Line DE placed on the Right Line AB the Point E shall fall upon B because DE is equal to AB also the Right Line DF shall fall upon AC because the Angle A is equal to D Moreover the Point F shall fall on the Point C because AC is equal to DF Therefore the Right Lines EF BC shall agree because they have the same terms and so consequently are equal wherefore the Triangle BAC is equal to DEF and the Angles B E as also the Angles C F do agree and are equal which was to be demonstrated USE Use 4. SVppose I was to measure the inaccessible Line AB I look from the Point C to the Point A and B then I measure the Angle C thus I place a Board or Table Horizontally and looking successively with a Ruler towards the Points A and B I draw two Lines making the Angle ACB then I measure the Lines AC and BC which I suppose to be accessible I turn about my Board or Table towards some other place in the Field placing it again Horizontally at the Point F and looking along those Lines I have drawn on my Table I make the Angle DEF equal to the Angle C I make also FD FE equal to CA CB Now according to this Proposition the Lines AB DE are equal wherefore whatsoever the Length of the Line DE is found to be the same is the measure of the inaccessible Line AB Another USE may be this Use 4. SVppose you were at a Billiard Table and you would strike a Ball B by reflection with another Ball A Admit CD be one Side of the Table now imagine a perpendicular Line BDE I take the Line DE equal to DB I say that if you aim and strike your Ball A directly towards the Point E the Ball A meeting the Side of the Table at F shall reflect from thence to B for in the Triangle BFD EFD the Side FD is common and the Sides BD DE equal the Angles BFD EFD equal by the Proposition the Angles AFC DFE being opposite are also equal as I shall demonstrate hereafter therefore the Angle of incidence AFC is equal to the Angle of reflection BFD and by consequence the reflection will be from AF to FB PROPOSITION V. THEOREM IN every Isosceles Triangle the Angles which are above the Base are equal as also those which are underneath Let ABC be an Isosceles Triangle viz that the side AB AC be equal I say that the Angle ABC ACB are equal as also the Angles GBC HCB which ly under
of 4 Foot and thrice in the Line of 6 the first to the second shall have the same Reason as 2 to 3. Irrational Reason is between Two Magnitudes of the same Species which are incommensurable that is to say that have not a common measure As the Reason of the Side of a Square to its Diagonal For there cannot be found any measure although never so little which will measure both precisely Four Magnitudes shall be in the same Reason or shall be Proportionals when the Reason of the first to the second shall be the same or like to that of the third to the fourth wherefore to speak properly Proportion is a similitude of Reason But one findeth it difficult to understand in what consisteth this similitude of Reason It is only to say that two habitudes or Relations be alike For Euclid hath not given a just Definition and which might have explained its Nature having contented himself to give us a mark by which we may know if Magnitude have the same Reason And the obscurity of this Definition hath made this Book difficult I will endeavour to supply this default 5. Euclide saith that Four Magnitudes have the same Reason when having taken the Equi-multiplices of the first and of the third and other Equimultiplices of the second and of the fourth whatever combination is made when the Multiplex of the first is greater than the Multiplex of the second the Multiplex of the third shall be also greater than the Multiplex of the fourth And when the Multiplex of the first is equal or less than the Multiplex of the second the Multiplex of the third is equal or less than the Multiplex of the fourth That then there is the same Reason between the first and second as there is between the third and fourth A B C D 2. 4. 3. 6. E F G H 10 8. 15 12. K L M N 8. 8. 12. 12. O P Q R 6. 16 9. 24 As if there were proposed four Magnitudes A B C D. Having taken the Equi-multiplexes of A and C which let be E and G quintuplex F and H double to B and D. In like manner taking K and M quadruple to A and C L and N double to B and D. Taking again O and Q triple to A and C P and R quadruple to B and D. Now because E being greater than F G is greater than H and K beng equal to L M is equal to N In fine O being lesser than P Q is lesser than R. Then A shall have the same Reason to B as C to D. I believe that Euclid ought to have Demonstrated this Proposition seeing it is so intangled that it cannot pass for a Maxim To explain well what Proportion is it is to say that four Magnitudes have the same Ratio although one may say in general that to that end the first must be alike part or a like whole in respect of the second as is the third compared to the fourth notwithstanding because this Definition doth not convene with the Reason of equality there must be given a more general and to make it intelligible it must be explained what is meant by a like Aliquot part Like Aliquot parts are those which are as many times in their whole as three in respect of nine two in respect of six are alike Aliquot parts because each are found three times in their whole The first quantity will have the same Reason to the second as the third hath to the fourth if the first contains as many times any Aliquot part of the second whatever as the 3d. contains alike Aliquot part of the 4th A B C D. as if A contains as many times a Hundreth a Thousandth a Millionth part of B as C contains a Hundreth a Thousandth a Millionth part of D and so of any other Aliquot parts imaginable there will be the same Reason of A to B as of C to D. To make this Definition yet clearer I will in the first place prove that if there be the same reason of A to B as there is of C to D A will contain as many times the Aliquot parts of B as C doth of D. And I will afterwards prove that if A contains as many times the Aliquot parts of B as C doth of D there will be the same Reason of A to B as of C to D. The first Point seemeth evident enough provided one doth conceive the terms for if A contains one Hundred and one times the tenth part of B and C only One Hundred times the tenth part of D the Magnitude A compared with B would be a greater whole than C compared with D so that it could not be compared after the same manner that is to say the habitude or Relation would not be the same The second point seemeth more difficult to wit whether if this propriety be so found the Reason shall be the same that is to say if AB contains as many times any Aliquot parts whatever of CD as E contains like Aliquot parts of F there shall be the same Reason of AB to CD as of E to F. For I will prove that if there were not the same Reason A would contain more times any Aliquot part of B than C containeth alike Aliquot parts of D which would be contrary to what we had supposed Demonstration Seeing there is the same Reason of AG to CD as of E to F AG will contain as many times KD an Aliquot part of CD as E would contain a like Aliquot part of F. Now AB contains KD once more than AG thence AB will contain once more KD an Aliquot part of CD than E doth contain a like Aliquot part of F which would be contrary to the supposition 6. There will be a greater Reason of the first quantity to the second than of the third to the fourth if the first contains more times any Aliquot part of the second than the third doth contain a like Aliquot part of the fourth As 101 hath a greater Reason to 10 than 200 to 20 because that 101 contains One Hundred and one times the Tenth part of 10 and 200 contains only One Hundred times the Tenth part of 20 which is 2. 7. The Magnitudes or quantities which are in the same Reason are called Proportionals 8. A Proportion or Analogie is a Similitude of Reason or habitude 9. A Proportion ought to have at least three terms For to the end there be similitude of Reason there must be two Reasons Now each Reason having two terms the antecedent and the consequent it seemeth there ought to be four as when we say that there is the same Reason of A to B as of C to D but because the consequent of the first Reason may be taken for antecedent in the second three terms may suffice as when I say that there is the same Reason of A to B as of B to C. 10. Magnitudes are in continued Proportion
to EFG Demonstration We have already demonstrated that there is a greater Reason of A B C to E F G than of the part BC to the part FG There shall thence be a greater Reason of A to E than of A B C to E F G by the 32d The end of the Fifth Book THE SIXTH BOOK OF Euclid's Elements THis Book explaineth and beginneth to apply perticular matters of the Doctrine of Proportions which the preceding Book Explaineth but in General It beginneth with the most easiest figures that is to say Triangles giving Rules to determine not only the Proportion of their Sides but also that of their capacity area or superficies Then it teacheth to find Proportional Lines and to augment or diminish any figure whatever according to a given Ratio It Demonstrateth the Rule of Three it applieth the forty seventh of the first to all sorts of Figures In fine it giveth us the most easie and most certain principles to conduct us in all sorts of Measuring The DEFINITIONS 1. Def. 1. Fig. I. Plate 6. RIght Lined Figures are like when that they have all their Angles equal and the sides which formeth those Angles Proportional Fig. I. As the Triangles ABC DEF shall be like if the Angles A and D B and E C and F are equal and if there be the same Ratio of AB to AC as of DE to DF and of AB to CB as of DE to EF. 2. Fig. II. Figures are reciprocal when they may be compared after such sort that the Antecedent of one Reason and the Consequent of the other be found in the same Figure That is when the Analogie beginneth in one figure and endeth in the same As if there be the same Reason of AB to CD as of DE to BF 3. Fig. III. A Line is divided into extream and mean Proportion or Reason when there is the same Reason of the whole Line to its greatest part as of its greatest part to its lesser part As if there be the same Reason of AB to AC as of AC to CB the Line AB shall be divided in the point C in extream and mean proportion 4. The Altitude or height of a figure is the Length of the Perpendicular drawn from the top thereof to its Base Fig. IV. As in the Triangles ABC EFG the Perpendicular EH AD whether it fall without or within the Triangles is their height or Altitude Triangles and Parallelograms whose heights are equal may be placed between the same Parallels for having placed their Bases on the same Line HC if the Perpendiculars DA HE are equal the Lines EA HC shall be parallel 5. A Reason is composed of several Reasons when the Quantities of those Reasons being Multiplied make a Third Reason It is to be taken notice of that a Reason at least the rational hath its name taken from some number which specifieth the Reason or habitude of the Antecedent of that Reason to its Consequent As when there is proposed Two Magnitudes or Quantities the one of Twelve Foot and the other of Six we say that the Reason of Twelve to Six is double In like manner when there is proposed Two Magnitudes Four and Twelve we would say it is a subtriple Reason and â…“ is the Denominator thereof which specifieth that there is the same Reason of Four to Twelve as of â…“ to One or as of One to Three One may call this Denominaotr the Quantity of the Reason Let there be proposed Three Terms Twelve Six Two The First Reason of Twelve to Six is double its Denominator is Two the Reason of Six to Two is Triple its Denominator is Three the Reason of Twelve to Two is compounded of the Reason of Twelve to Six and of that of Six to Two to have thâ—… Denominator or the Reason of Twelve to Two which is compounded of Double and of Triple multiply Three by Two and you shall have Six thence the Reason of Twelve to Two is Sextuple This is that which Mathematicians understand by composition of Reason although it ought to be called Multiplication of Reason PROPOSITION I. THEOREM PArallelograms and Triangles of equal height have the same Reason as their Bases Plate VI. Let there be proposed the Triangles ACG DEM of equal height in such sort that they may be placed between the same Parallels AD GM I say that there shall be the same Reason of the Base GC to the Base EM as of the Triangle AGC to the Triangle DEM Let the Base EM be divided into as many equal parts as you please and let there be drawn to each Division the Lines DF DH c. Let also the Line GC be divided into parts equal to those of EM and let be drawn Lines to each division from the top A All those little Triangles which are made in the Two great Triangles are between the same Parallels and they have equal Bases they are thence equal by the 28th Demonstration The Base GC contains as many Aliquot parts of the Line EM as could be found parts equal to EF Now as many times as there are in the Base GC parts equal to EF so many times the Triangle AGC containeth the little Triangles equal to those which are in the Triangle DEM which being equal among themselves are its Aliquot parts thence as many times as the Base GC containeth Aliquot parts of EM so many times the Triangle AGC containeth Aliquot parts of the Triangle DEM which will happen in all manner of Divisions There is thence the same Reason of the Base GC to the Base EM as of the Triangle AGC to the Triangle DEM Coroll Parallelograms drawn on the same Bases and that are between the same Parallel Lines are double to the Triangle by the 41st they are thence in the same Reason as Triangles that is to say in the same Reason as their Bases USE Fig. I. THis Proposition is not only necessary to Demonstrate those which follow but we may make use thereof in Dividing of Land Let there be proposed a Trapezium ABCD which hath its sides AD BC parallel and admit one would cut of a third part Let CE be made equal to AD and BG the third part of BE. Draw AG. I say that the Triangle ABG is the Third of the Trapezium ABCD. Demonstration The Triangles ADF FCE are equiangled because of the Parallel AD CE and they have their Sides AD CE equal They are thence equal by the 26th of the 1st and by consequence the Triangle ABE is equal to the Trapezium Now the Triangle ABG is the third part of the Triangle ABE by the preceding Proposition thence the Triangle ABG is the third part of the Trapezium ABCD. PROPOSITION II. THEOREM A Line being drawn in a Triangle Parallel to its Base divideth its sides proportionally and if a Line divideth proportionally the sides of a Triangle it shall be parallel to its Base If in the
the double Area of the Triangle I make use thereof in several other Propositions as in the seventh PROPOSITION XIV PROBLEM TO describe a Square equal to a right lined Figure given To make a Square equal to a Right lined Figure A make by the 45th of the 1st a Rectangle BC DE equal to the Right lined Figure A if its sides CD DC were equal we should have already our desire if they be unequal continue the Line BC until CF be equal to CD and dividing the Line BF in the middle in the Point G describe the Semicircle FHB then continue DC to H the Square of the Line CH is equal to the Right lined Figure A draw the Line GH Demonstration The Line BF is equally divided in G and unequally in C thence by the 5th the Rectangle comprehended under BC CF or CD that is to say the Rectangle BD with the Square CG is equal to the Square of GB or to its Equal GH Now by the 47th of the 1st the Square of GH is equal to the Square of CH CG therefore the Rectangle BD and the Square of CG is equal to the Squares of CG and of CH and taking away the Square of CG which is common to both the Rectangle BD or the Right lined Figure A is equal to the Square of CH. USE THis Proposition serveth in the first place to reduce into a Square any Right lined Figure whatever and whereas a Square is the first Measure of all Superficies because its Length and Breadth is equal we measure by this means all right lined Figures In the second place this Proposition teacheth us to find a mean Proportion between two given Lines as we shall see in the Thirteenth Proposition of the Sixth Book This Proposition may also serve to square curve lined Figures and even Circles themselves for any crooked or curve lined Figure may to sence be reduced to a Right lined Figure as if we inscribe in a Circle a Polygon having a thousand sides it shall not be sensibly different from a Circle and reducing the Polygone into a Square we square nearly the Circle THE THIRD BOOK OF Euclid's Elements THis Third Book explaineth the Propriety of a Circle and compareth the divers Lines which may be drawn within and without its Circumference It farther considereth the Circumstances of Circles which cut each other or which touch a streight Line and the different Angles which are made as well those in their Centers as in their Circumferences In fine it giveth the first Principles for establishing the Practice of Geometry by the which we make use and that very commodiously of a Circle in almost all Treatises in the Mathematicks DEFINITIONS 1. Def. of the 3 Book Those Circles are equal whose Diameters or Semidiameters are equal 2. Fig. 1. A Line toucheth a Circle when meeting with the Circumference thereof it cutteth not the same as the Line AB 3. Fig. 2. Circles touch each other when Meeting they cut not each other as the Circles AB and C. 4. Fig. 3. Right Lines in a Circle are equally distant from the Center when Perpendiculars drawn from the Center to those Lines are equal As if the Lines EF EG being Perpendiculars to the Lines AB CD are equal AB CD shall be equally distant from the Center because the Distance ought always to be taken or measured by Perpendicular Lines 5. Fig. 4. A Segment of a Circle is a Figure terminated on the one side by a streight Line and on the other by the Circumference of a Circle as LON LMN 6. The Angle of a Segment is an Angle which the Circumference maketh with a streight Line as the Angle OLN LMN 7. Fig. 5. An Angle is said to be in a Segment of a Circle when the Lines which form the same are therein as the Angle FGH is in the Segment FGH 8. Fig 6. An Angle is upon that Arch to which it is opposite or to which it serveth for a Base as the Angle FGH is upon the Arch FIH which may be said to be its Base 9. Fig. 6. A Sector is a Figure comprehended under two Semidiameters and under the Arch which serveth to them for a Base as the Figure FIGH PROPOSITION I. PROBLEM To find the Center of a Circle IF you would find the Center of the Circle AEBD draw the Line AB and divide the same in the middle in the Point C at which Point erect a Perpendicular ED which you shall divide also equally in the Point F. This Point F shall be the Center of the Circle for if it be not imagine if you please that the Point G is the Center draw the Lines GA GB GC Demonstration If the Point G were the Center the Triangles GAC GBC would have the sides GA GB equal by the definition of a Circle AC CB are equal to the Line AB having been divided in the middle in the Point C. And CG being common the Angles GCB GCA would then be equal by the 8th of the 1st and CG would be then a Perpendicular and not CD which would be contrary to the Hypothesis Therefore the Center cannot be out of the Line CD I further add that it must be in the Point F which divideth the same into two equal Parts otherwise the Lines drawn from the Center to the Circumference would not be equal Corollary The Center of a Circle is in a Line which divideth another Line in the middle and that perpendicularly USE THis first Proposition is necessary to demonstrate those which follow PROPOSITION II. THEOREM A Streight Line drawn from one point of the Circumference of a Circle to another shall fall within the same Let there be drawn from the Point B in the Circumference a Line to the Point C. I say that it shall fall wholly within the Circle To prove that it cannot fall without the Circle as BVC having found the Center thereof which is A draw the Lines AB AC AV. Demonstration The Sides AB AC of the Triangle ABC are equal whence by the 5th of the 1st the Angles ABC ACB are equal And seeing the Angle AVC is exteriour in respect of the Triangle AVB it is greater than ABC by the 16th of the 1st it shall be also greater than the Angle ACB Thence by the 19th of the 1st in the Triangle ACV the side AC opposite to the greatest Angle AVC is greater than AV and by consequence AV cannot reach the circumference of the Circle seeing it is shorter than AC which doth but reach the same wherefore the Point V is within the Circle the same may be proved of any Point in the Line AB and therefore the whole Line AB falls within the Circle USE IT is on this Proposition that are grounded those which demonstrate that a Circle toucheth a streight Line but only in one Point for if the Line should touch two Points of its Circumference it would be then drawn from one