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ground_n line_n right_a superficies_n 3,973 5 16.1798 5 true
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ID Title Author Corrected Date of Publication (TCP Date of Publication) STC Words Pages
B08245 The accomplish'd sea-mans delight containing : 1. The great military of nature demonstrated by art ... 2. The closset of magnetical miracles unlocked ... 3. Directions for sea-men in distress of weather ... 4. The resolver of curiossities being a profitable discourse of local ... 1686 (1686) Wing A167A; ESTC R215626 100,294 169

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should make an Angle equal to the Angle of Contigence To this I answer that although a greater Angle be granted yet not a less for if the less should be granted as well as the greater we should likewise have an equal For Example consider the circle A C with the Line A B doth touch in the point A. The Angle of contingence shall be A B C then let the circle inwardly described A E touching the circle A C in the point A for to one only point it shall touch the circle A C by twelve of the third And so the Line A B shall touch the circle A E by common Science by the which the Angle B A E shall be greater than the Angle B A C. Likewise also if the circle A D shall be described outwardly the Angle B A D should be less than the Angle B A C. And consequently by the same order whereby we make the greater or the less we shall also constitute equal which is the less Wherefore it followeth that it may be done contrary to that which Aristotle saith and for the same sentence of Aristotle some have thought that it is impossible for any of the figures of crooked Lines should be frund equal to any figure of a right Line or the contrary the which to be impossible I will demonstrate For Example let be given a Trigon I mean also of all figures of right Lines for as much as they shall be divisible into Triangles as appeareth by the thirty-second of the first And of these Triangles we shall constitute a superficial Line of equidistant sides by forty-four of the first Taken as often as need shall be which duplicate by the helps of thirty-six of the first and afterwards a Diameter in it then the healf of the superficies shall have an equal Triangle of the taken superficies of the forty-one of the first or by the taken right Line by the first conception I will constitute a superficial of two crooked Lines continued equal unto it I will divide the first basis or ground A C by equal spaces into points H by ten of the first and I draw B H which also I draw forth untill H K by double to B H by 3 of the first assumpted Then to the half of H K. Thus is I I direct C I and AI I joyn thereto also AK and C K by right Lines then by the first of the sixt these Triangles shall be all equal to themselves After this I will constitute a superficial of equidistant sides and of right Angles upon whatsoever Line which superficies shall be equal to the Poligonie ABCK by 44 of the first assumpted as often as shall be needful that superficies is made G D. But in the which I draw the Diameter F E so that by 41 of the whole Trigon F G E shall be the half of the whole superficies and by common science equal to the Tigon B K C. and Triplus to the Trigon B H C. Now I divide F G by equal in the point M by 10 of the first So do I also of the Line M L dividing it by equal in the point N by the aforesaid 10 of the first Afterwards by 44 of the first twice assumpted of equidistant sides I make a superficies of right Angles upon the Line M N equal to the quadrature of the Line F M traverse or overthwart and N D right I constitute a Parabol of a right Angle that it may be of less labour for this Example may suffice of by 52 of the first of Apolonius Pergeus the Termining Line of which Parabol shall pass by the points of F N and G by the same and by 33 of the same F E shall touch the Parabol at the point F. And afterwards when the Trigon F E G shall be triplus to the Trigon B H C as we have shewed before but also the portion of F N G Triplus by the 17 of Archimedes de quadratura Parabolae Wherefore the portion F N G shall be equal to the Trigon H B C by the first conception in Euclide added by Companus Furthermore I draw E G untill by the third of the first G R. Equal G R. I draw forth also F R and LMO then by the fourth of the first Triangle FGE shall be of equal sides and also of equal Angles to the Triangle F G R. Furthermore D M is equidistant G R by common science and by R G of the first the Angle F D M equal to the Angle F R G and the Angle of F R G equal to the Angle of F M D and whereas the Angle of F R G is common to either of them then by the four of the fix the same or all one shall be the proportion of R G to DM as is of G F to M F. But as is G F to M F so is G F to M L. Wherefore by N of the first G F hath it self so to M F as G R to DM But by the 16 of the same ML to DM hath it self and GF to GR. Wherefore ML equal MQ which MQ I divide by equal in the the point X by 10 of the first and will do as before Then by the reasons aforesaid of the same the portion of FXG shall be equal to the Trigon ABH and the whole superf●ices FGNX shall be equal to the whole Trigon ABC which is proposed The contrary appeareth thus Let be grated a superficies contained of two parallel lines as FNG and F X G proposing for Example to find a superficial of right lines Trianguler equal to the granted superficies I draw first F G. Then afterwards by 44 of the second of Apolonius Pergeus I find the Diameter of the Parabol FNG which is MN which I draw to ML to be equal MN Then I draw F L which shall touch the Parabol of F N G in the point F by 33 of the first of the same then from the point G I draw a line G E equidistant from the Diameter M N L by 31 of the first of Euclide which I draw untill it joyn together with F L the which doubtless shall be done by the second of the first of Vitellio The Point of the Concourse or joining together is E then I divide F into three equal portions by the 11th of the sixth of Euclide in the points S T which points I join with the point G. by the Lines of F G and G R. Now shall there be three Angles all equal to themselves by the 38 of Euclide After this I constitute a Trigon B H C equal to the Trigon F S G. by this meanes I draw forth H C to the Equality of G S by the Equality of the 4th of the first of Euclide Then at the point H I design an Angle B H C equal to the Angle of F S G. by the 23d of the first of Euclide and by 3. of the first of the same I draw H B until it be equal F S. Afterwards I