And because GH is equal to EL, C, but GH is equal to MN, and MN is therefore equal to EL, C. Take away EL, which is common. Therefore the remaining gnomon voy is equal to c. And because AE is equal to EB, AN is also equal to NB, that is to Lo. Add Ex, which is common; therefore the whole AX is equal to the gnomon vxY. But the gnomon vxY is equal to c; and Ax is therefore equal to c. Therefore to the given right line AB, a parallelogram c has been applied, exceeding by the parallelogram Po similar to D, because op is similar to D.b Q. E. F. PROPOSITION XXX. PROBLEM. To cut a given finite right line in extreme and mean ratio. b 24.6. Let AB be a given finite right line; it is required to cut the given right line AB in extreme and mean ratio. Describe upon Aв the square BC, and apply to AC the 46. 1. parallelogram CD equal to BC, exceeding by the figure AD similar to Bc.b But BC is a square, hence AD is also a square. And because BC is F H A B E D equal to CD, take away CE, which is Otherwise. Let AB be a given right line, it is required to cut AB in extreme and mean ratio. Divide AB in c, so that the rectangle under AB, BC, may be equal to the square of Ac.c b 29. 6. e 11. 2. d 17, 6. 4. 6. And because the rectangle under AB, BC, is equal to the square of ca; hence A C B it is as AB to AC so is AC to CB. Therefore AB has been cut in c in extreme and mean ratio. Deductions. Q. E. D. 1. A given right line being cut in extreme and mean ratio, if from the greater segment the less be taken, the greater segment also will thus be cut in extreme and mean ratio; and if a right line, equal to the greater segment, be added to the given line, the line, which is made up of the given line and this segment, is also cut in extreme and mean ratio. 2. Upon a given right line as an hypothenuse to describe a right-angled triangle, which shall have its three sides continual proportionals. PROPOSITION Xxxi. THEOREM. In right-angled triangles, the figure described upon the side subtending the right angle is equal to the figures, similar and similarly described upon the sides, containing the right angle. Let ABC be a right-angled triangle, having the right angle BAC; then the figure described upon BC is equal to the similar and similarly described figures upon BA, Draw the perpendicular AD. AC. C And because in the right-angled triangle BAC, a perpendicular AD is drawn upon the base BC from the right angle at A; therefore the triangles ABD, ADC, are BD similar to the whole ABC, and to one another. And because ABC is similar to ABD, therefore it is as CB to BA So is BA to BD. And because there are three proportionals, it is as the first to the third so is the figure described upon the first to the similar and similarly described figure upon the second; hence as CB to BD so is the figure upon cв to the similar and similarly described figure upon BA. For the same reason, as BC is to CD so is the figure upon BC to the similar and similarly described figure upon CA; wherefore also as Bc is to BD, DC, so is the figure upon BC to those upon CA, BA. But BC is equal to BD, DC; therefore also the figure upon BC is equal to the similar and similarly described figures upon BA, AC. Therefore in right-angled triangles, &c. Q. E. D. PROPOSITION XXXII. THEOREM. If two triangles having two sides of the one proportional to two sides of the other, be joined at one angle, so that their homologous sides be parallel; the remaining sides of those triangles shall be in one right line. Let ABC, CDE, be two triangles, having the two sides BA, AC, proportional to the two sides CD, DE, viz. as a B to AC so is DC to DE, also AB parallel to DC, and ac to DE; then BC, CE, are in one right line. B C E a 29. 1. For because AB is parallel to DC, and AC falls upon them, the alternate angles, BAC, ACD, are equal to one another. For the same reason, also, CDE is equal to ACD; wherefore also BAC is equal to CDE. And because there are two triangles ABC, DCE, having the angle at a equal to the angle at D, and the sides about the equal angles proportionals, viz. as BA to AC so is CD to DE. Therefore the triangle ABC is equiangular to the triangle CDE; hence the angle ABC is equal to DCE. 6. 6. But it has been shown that ACD is equal to BAC: hence the whole ACE is equal to the two ABC, BAC; add the common angle ACB; therefore the angles BAC, ABC, BCA, are equal to two right angles, and ACE, ACB. To any 32. 1. right line AC, and at any point c, two right lines BC, CE, not placed at the same parts, make the adjacent angles ACE, ACB, equal to two right angles; hence BC is in the same right line with CE. b b d 14. 1. N a 27.3. b 27.3. PROPOSITION XXXIII. THEOREM. In equal circles, angles have the same ratio as the circumferences on which they stand, whether they be at the centres, or at the circumferences; and so also are the sectors, as being at the centres. Let ABC, DEF, be equal circles, and BGC, EHF, at their centres G, H, also BAC, EDF, at the circumferences; then as the circumference BC is to the circumference EF so is the angle BGC to EHF, and BAC to EDF, also the sector GBC to the sector HEF. Take any number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN, each equal to EF, and join GK, GL, HM, HN. And because the A to one another, K D D M E F If and the angles BGC, CGK, KGL, equal to one another. Whatsoever multiple the circumference BL is of BC, the same multiple is the angle BGL of BGC. For the same reason, whatsoever multiple the circumference EN is of EF, the same multiply is the angle EHN of EHF. therefore the circumference BL be equal to the circumference EN, the angle BGL is also equal to EHN; and if the circumference BL be greater than the circumference EN, the angle BGL is greater than the angle EHN; and if less, less; hence there being four magnitudes, the two circumferences BC, EF, also the two angles BGC, EHF, of the circumference BC and of the angle BGC are taken any equimultiples whatsoever, viz. the circumference BL and the angle, BGL also of the circumference EF and angle EHF, viz. the circumference EN and the angle EHN; and it has been shown if the circumference BL exceed the circumference EN, the angle BGL will exceed the angle EHN; if equal, equal; and if less, less; therefore it is as the circumference c 3 Def. 5. BC to EF so is the angle BGC to EHF. But as the angle BGC to EHF So is BAC to EDF, for each is double of each; and hence as the circumference BC is to the circumference EF so is the angle BGC to EHF and BAC to EDF. Therefore in equal circles, angles have the same ratio as the circumferences on which they stand, whether at the centres or at the circumferences. Q. E. D. Again, as the circumference BC is to the circumference EF so is the sector GBC to the sector HEF. A D H For join BC, CK, and take any points x, o, in the circumference, BC,CK,and join вX, XC, co, ok. And because the two BG, GC, are equal to the two CG GK, and they comprehend equal angles, the base BC is K L M E F also equal to CK; therefore the triangle BGC is equal to the triangle GCK. And because the circumference BC is equal to the circumference сK, also the remaining circumference of the whole circle is equal to the remaining circumference of the whole circle; wherefore also the angle BXC is equal to the angle coк; hence the segment вXC is similar to the segment coK; and they are upon equal right lines вс, ск. But similar segments of circles upon equal right lines are equal to one another; hence the segment BXC is equal to the segment coк. But the triangle BGC is equal to the triangle GCK; and therefore the whole sector GBC is equal to the whole sector GCK. For the same reason the sector GKL is equal to each of them GKC, GCB; hence the three sectors GBC, GCK, GKL, are equal to one another. For the same reason also the sectors HEF, HFM, HMN, are equal to one another; therefore whatsoever multiple the circumference BL is of the circumference BC, the same multiple is the sector GBL of the sector GBC. For the same reason also whatsoever multiple the circumference EN is of the circumference EF, the same multiple is the sector HEN of the sector HEF; if therefore the circumference BL is equal to the circumference EN, the sector BGL is also equal to the sector HEN; and if the circumference BL exceed the circumference EN, the sector GBL will also exceed the sector HEN; and if less, less. Hence there being four magnitudes, the two circumferences BC, EF, also the two sectors GBC, HEF, and any equimultiples of the circumference BC and of the sector GBC, are taken, viz. the circumference BL, and the sector GBL; also any equimultiples of the circumference EF and sector HEF, viz. the circumference EN, and sector HEN. And it has been shown that if the circumference BL exceed the circumference EN, the |