Selected quad for the lemma: reason_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
reason_n	angle_n	equal_a	line_n	4,117	5	11.1250	5	true

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A47139	An essay for the discovery of some new geometrical problems (judged by some learned men, impracticable) concerning angular sections, beginning with the geometrical trisection of any right lined angle, by plain geometry of right lines and arches of circles, with rule and compass only, with out all conick sections, and cubick æquations. Whether the following praxis, and apparent demonstration thereof doth not only make it practicable, but easie to the understanding of a tiro, who but understands a little in true geometrical learning. Which layeth a foundation of a plain method how to sect any angle into any other number of parts required, even as 4. 6. 8. 10; or uneven, as 5. 7. 9. 11. &c. As also to divide a circle into any number even, or uneven of equal parts. All which have great uses in the improvement of the mathematical sciences, some of which are here specified. Proposed and submitted to the impartial tryal and examination of the right reason of such artises, to whose hands it may come. By G.K.	Keith, George, 1639?-1716.	1697 (1697)	Wing K160; ESTC R221663	9,043	17

AN ESSAY For the Discovery of Some NEW Geometrical Problems Judged by some Learned Men Impracticable Concerning ANGULAR SECTIONS Beginning with the GEOMETRICAL TRISECTION of any Right Lined Angle by Plain Geometry of Right Lines and Arches of Circles With RVLE and COMPASS only without all Conick Sections and Cubick Aequations Whether the following Praxis and apparent demonstration thereof doth not only make it Practicable but easie to the Understanding of a Tiro who but understands a little in true Geometrical Learning Which layeth a Foundation of a Plain Method how to Sect any Angle into any other Number of Parts required Even as 4. 6. 8. 10 or Vneven as 5. 7. 9. 11. c. As also to divide a Circle into any number Even or Vneven of equal parts All which have great Uses in the Improvement of the Mathematical Sciences some of which are here specified Proposed and Submitted to the Impartial Tryal and Examination of the Right Reason of such Artists to whose Hands it may come By G. K. London Printed 1697. And to be Sold by the Author at his House in Pudding-Lane at the Sign of he Golden-Ball near the Monument And by B. Aylmer at the Three Pigeons over against the Royal-Exchange Some New Geometrical Problems c. ALthough the Trisection of a right lined Angle and also its Section into any parts required to a true Mathematical exactness is denyed to be practicable by plain Geometry of right Lines and Arches of Circles with Rule and Compass only without all Conick Sections and Algebra Equations by some learned Artists yet others as learned are not so positive but acknowledge the Practise of it is not as yet discovered among whom is the learned Is Barrow who hath writ thus in Corol. ad 9. 1. elem Eucl. Methodus vero regula Circino angulos secandi in aequales quotcunque hactenus Geometras latuit and in Schol. ad 16. 4. Elem. Coeterum divisio circumferentiae in partes datas etiamnum desideratur The Praxis of the Trisection of any given Angle 1. Divide the Cord BC into 3 equal Parts as BE = EE = EC by 10. 6. elem Euclid and draw the 2 Perpendiculars EF. EF. 2. With the extent of ⅓ of the Cord as EC measure on the Arch from C to G and draw the Line or Cord GC = EC 3. With another Radius less than AC viz. AK draw a second arch as KL until it cut the perpendicular EF and let the Radius AK be so long that the extent of the Cord betwixt K and L be somewhat longer than EC and with the same extent EC measure on this second arch KR = EC 4. From G to R draw a straight Line by post 1. 1. elem and produce the same by post 2. until it cut the Perpendiculars at I and O. 5. Draw the straight lines AI and AO and extend them to H on both sides so that the one line shall be AIH and the other AOH which two Lines shall sect the given angle BAC into three parts or angles viz. BAH HAH HAC The Question or Problem to be resolved is whether these three angles are not equal and consequently that the angle BAC is trisected into equal angles The apparent demonstration that is here proposed to prove it followeth In order to the apparent following demonstration by way of Preparation 1. Draw the line OI which shall be parallel to EE by 28. 1. elem and extend it on both sides from I to N on the one side and from O to N on the other making IN = ON = EC = OI 2. With Radius IN and ON describe the arches on both sides NM until these arches cut the lines AC and AB so shall the lines or cords drawn betwixt I and M and O and M viz. IM and OM = EC 3. Make the angle IMP = angle MIN by 23. 1. elem and OMP = MON and extend the line MP until it cut the line AI at P and AO at P. 4. Draw the line OP parallel to IM until it cut MP whence the perfect Rhombus OMPI shall be formed having the opposit side paralel and equal for PM by construction is parallel with OI and OP with IM and that the Line OP drawn parallel to IM can cut the Line PM no where but at P on the line AI is proved thus Suppose it cut the line PM any where else than on the line AI on either side it should make PM either longer or shorter than its opposit side for the arch drawn by the radius MI and cutting the line AI at P proveth MP = MI by def 15. 1. elem as also OP should be longer or shorter than IM as the like arch conceived to be drawn by radius OI and cutting AI at P proveth OI = OP by the same def 15. 1. elem but both these Consequences are absurd making the opposit sides of a Paralellogram to be unequal contrary to 34. 1. elem Because the Figure OPMI is proved to be a perfect Rhombus having all the sides equal whereof the line PI is the diagonal it being proved that OP and PM terminate on AI therefore the angle OIP = OIA = PIM = AIM by 8. and 4. 1. elem Lastly these two Triangles OAI and IAM shall be according to the 4th prop. 1. elem and so shall OAI and OAM for OI = IM as above proved and AI is common to both and the Angle OIA = AIM as is above proved therefore by 4. 1. elem AM = AO = AI and consequently by 8. 1. the angles IAM = OAI = OAM therefore the given angle BAC is Geometrically trisected by the Lines AO and AI extended to H. Q. E. D. Here Note let the Radius be ever so much changed betwixt M and C taking it at any extent from A and let ever so many Concentrick Arches be drawn from the Center A betwixt IM and GC their Cords terminating on the streight Lines MC on the one side and IG on the other shall all be equal one to another and to the Cord GC = ●BC as the demonstration above given proveth for the reason that proveth the Cords of any 3 Concentrick Arches terminating on two streight Lines to be equal within the limits MI and GC being a Trapezia proveth any other 3 to be equal within the same limits But if we draw any Concentrick Arches without the Trapezia IGCM as with a less Radius than AM or with a greater Radius than AC the Case is altered and the equal Cords will not terminate on the streight Line IG but diviate or depart therefrom as both true Reason doth prove and even ocular inspection doth shew for though the Eye is not able to judge of a straight Line yet when a Line is visibly apparent to make an Angle with another Line these two Lines cannot be a straight Line I call the Figure IMCG a Trapezia for it is not any Paralellogram because the

BFC and AO to X. Again with the extent AI on the center I describe the circle ASHT. and with the same Radius describe the arch KLON Again draw IH parallel to AK extending IH to the Circumference of the Circle ASHT. Again draw AS parallel and equal to OI which shall cut the circle at S because OI = AS cuts the arch KION having the same Radius Again from the point S on the circumference draw the line SH parallel to AI which shall necessarily cut the line IH on the circumference at the point H because by construction IH is parallel to AK and KH to AI which makes the Parallogram AKHI whose opposit sides are equal but if SH and IH did meet any where else than on the circumference as either within or without it the opposit sides of the Parallogram should not be equal contrary to 34. 1. Eucl. elem Again draw the line SI and extend the line OI from I to K until it cut SH at K which shall form another Parallogram ASKI which being divided by the Diagonal line SI shall give two equal Isosceies Triangles AIS = OAI ISK. The Demonstration In the Isoceles triangle ISH whose sides IS = IH by 15. def 1. elem the angle IHS = ISH by 5. 1. elem and IHS = IAK by 8. 1. elem therefore IAK = ISH but ISH = AIS = OAI being alternate Angles betwixt Parallels by 29. 1. elem and therefore lasty OAI = IAK and by the like method OAI is proved = OAN drawing the like parallel Lines on the other side of the Triangle OAI and therefore the angle BAC is trisected by 3 equal angles NAO = OAI = IAK = BAX = XAD = DAC Q. E. D. The Praxis of the Section of a Right Lined Angle into Six equal parts BEcause of the affinity of the Figure of the Trisection with the Figure of the Sextisection I shall begin with the Sextisection and then proceed to the Quinquisection In the 3d figure let the given angle be BAC as above whose Cord is BC and its arch BEFGC 1. Divide the Cord BC into 6 equal parts which Divisions are marked with ***** and on the middle division draw the line ADI to the arch BEFGC 2. With the extent of one of the Divisions on the given arch measure from C to F and with the same extent from F to E and from E to G marking the points E. F. G. and with the same extent measure from B to E from E to F and from F to G marking the points EFG 3. With a less Radius than AC as AK draw a second Arch as KLHLK which Radius must be so long that the extent of the 1 6th of the Cord thrice repeated may fall short of reaching the line ADI 4. With the same extent on this 2d arch measure from AC and from AB to K and from K to L and from L to H by a threefold repitition of the same extent 5. From the point G on the first arch to the point H on the second arch draw a straight line and produce it until it cut the line ADI at the point I. 6. With the Radius AI draw a 3d. concentrick Arch betwixt the lines AB and AC I say the extent of the 16th part of the cord BC = MC being 6 times repeated shall cut the said 3d arch into six Equal parts to every one of which straight lines drawn from the Center A and produced to the outmost arch shall divide the given angle BAC into six equal parts orangles marked with 1. 2. 3. 4. 5. 6. In order to the demonstration of the Praxis of this Section of an angle into 6 equal parts on the point 5 making it a center with Radius 5A describe the circle AOPQRS 2. From the point 5 draw the line 5R parallel to AC 3. From R draw the line RT parallel to 5A 4. Draw the line AT from A to T 5. Draw the line T5 and draw 5V parallel to AT The Demonstration In the triangle T5R being an Isosceles the angle 5TR = 5RT by 5. 1. elem and 5RT = 5R6 5A6 therefore 5T6 = 5A6 Again 5 T6 = 5TV A5T and A5T = 5A4 by 34.1 and therefore lastly 5 A4 = 5 AV. and as these 2 angles are proved equal by the like reasoning all the other angles can be proved equal by changing the center of the Circle from 5 to 4 and from 4 to 3 and from 3 to 2 and from 2 to 1. This demonstration having such affinity with the former of Trisection and which can as easily be demonstrated both ways as that of Trisection I shall not enlarge upon it for he who understands and assents to the verity of the former by the same evidence will assent to the later The Praxis of the Quinquisection 1. ACcording to the former method divide the cord of the given angle BE into 5 = parts 2. On the middle part marked with CD draw the lines CH and DH cutting the cord at right angles and parallel one to another 3. On the arch of the given angle BFG take the â…• of the cord = CD and set it from E to F and from F to G also do the like from B to F and from F to G. 4. Draw a second arch with a less radius as AK so as the radius may be so long that the extent of the â…• of the cord twice taken on that second arch may not reach to the perpendicular DH 5. With the same extent twice taken as from K to * from * to * measure on the second arch to the second * 6. From G to * draw a straight line until it cut the perpendicular line DH at H lastly with the radius AH describe the third arch IHHI and draw the lines AH AH which shall make the angle HAH = â…•BAE The Demonstration of this or any other Section being after the same method with the former as also the Praxis it were superfluous to enlarge upon it or to add any new Problems showing how to sect any angle into any other parts given as 7. 8. 9. 10. 11. c. for he who understands by the foregoing Method and Praxis to sect any angle into 3. 5. 6. as is above shewed will by the like Method and Praxis be able to sect any angle into 7. 8. 9. 10. c. equal parts and also to demonstrate the same Wherefore in the next place I shall proceed to shew how a Semicircle may be sected into any number of equal parts even as 4. 6. 8. 10 c. or uneven as 5. 7. 9. 11. The Praxis of a Section of a Semicircle into 9 equal parts LEt the Semicircle given be BRSC whose diameter is BC. and whose center is A. 1. Divide the diameter BC into 9 equal parts 2. From the center A set off AO and AN making NO = 1 9 of the diameter and draw the perpendiculars NR OS 3. With the extent NO