D angle EHN of the angle EHF. But if the arc BL be equal to the arc EN, the angle BGL is also equal (III. 27) to the angle EHN; or if the arc BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less. There being then four magnitudes, the two arcs BC, EF, and the two angles BGC, EHF; and of the arc BC, and of the B H N K E M F angle BGC have been taken any equimultiples whatever, viz., the arc BL, and the angle BGL; and of the arc EF, and of the angle EHF, any equimultiples whatever, viz., the arc EN, and the angle EHN. And it has been proved, that if the arc BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if less, less. As, therefore, the arc BC to the arc EF, so (V. Def. 5) is the angle BGC to the angle EHF. But as the angle BGC is to the angle EHF, so is (V. 15) the angle BAC to the angle EDF, for each is double of each (III. 20). Therefore, as the circumfer ence BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. A D Also, as the arc BC to EF, so is the sector BGC to the sector EHF. Join BC, CK, and in the arcs BC, CK take any points X, O, and join BX, XC, CO, OK. Then, because in the triangles GBC, GCK, the two sides BG, GC are equal to the two CG, GK, and also contain equal angles; the base BC is equal (I. 4) to the base CK, and the triangle GBC to the triangle GCK. And because the arc BC is equal to the arc CK, the remaining part B of the whole circumference of the circle ABC X K H N M F is equal to the remaining part of the whole circumference of the same circle. Wherefore, the angle BXC is equal to the angle COK (III. 27); and the segment BXC is therefore similar to the segment COK (III. Def. 9); and they are upon equal straight lines BC, CK. But similar segments of circles upon equal straight lines are equal (III. 24) to one another. Therefore, the segment BXC is equal to the segment COK. And the triangle BGC, is equal to the triangle CGK; therefore the whole, the sector BGC, is equal to the whole, the sector CGK. For the same reason, the sector KGL is equal to each of the sectors BGC, CGK; and in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another. Therefore, what multiple soever the arc BL is of the arc BC, the same multiple is the sector BGL of the sector BGC. For the same reason, whatever multiple the arc EN is of EF, the same multiple is the sector EHN of the sector EHF. Now, if the arc BL be equal to EN, the sector BGL is equal to the sector EHN; and if the arc BL be greater than EN, the sector BGL is greater than the sector EHN; and if less, less. Since, then, there are four magnitudes, the two arcs BC, EF, and the two sectors BGC, EHF; and of the arc BC, and sector BGC, the arc BL and the sector BGL are any equimultiples whatever; and of the arc EF, and sector EHF, the arc EN and sector EHN are any equimultiples whatever; and it has been proved, that if the arc BL be greater than EN, the sector BGL is greater than the sector EHN; if equal, equal; and if less, less; therefore (V. Def. 5), as the arc BC is to the arc EF, so is the sector BGC to the sector EHF. Wherefore, in equal circles, &c. Q. E. D. If an angle of a triangle be bisected by a straight line, which likewise cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle. B C D Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; the rectangle BA.AC is equal to the rectangle BD.DC, together with the square of AD. Describe the circle (IV. 5) ACB about the triangle, and produce AD to the circumference in E, and join EC. Then, because the angle BAD is equal to the angle CAE, and the angle ABD to the angle (III. 21) AEC, for they are in the same segment; the triangles ABD, AEC are equiangular to one another. Therefore, BA: AD: : EA: AC (VI. 4) and, consequently, BA.AC= (VI. 16) AD.AE=ED.DA+DA2 (II. 3). But ED.DA=BD.DC (III. 35), therefore, BA.AC=BD.DC÷DA2. Wherefore, if an angle, &c. Q. E. D. PROP. C. THEOR. E If from any angle of a triangle a straight line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular, and the diameter of the circle described about the triangle. Let ABC be a triangle, and AD the perpendicular from the angle A to the base BC; the rectangle BA.AC is equal to the rectangle contained by AD and the diameter of the circle described about the triangle. Describe (IV. 5) the circle ACB about the triangle, and draw its diameter AE, and join EC. Because the right angle BDA is equal (III. 31) to the angle ECA in a semicircle, and the angle ABD to the angle AEC in the same B E segment (III. 21); the triangles ABD, AEC are equiangular. Therefore (VI. 4), BA to AD, so is EA to AC; and, consequently, the rectangle BA.AC is equal (VI. 16) to the rectangle ÉA.AD. Therefore, if from an angle, &c. Q. É. D. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides.* Let ABCD be any quadrilateral inscribed in a circle, and let AC, BD be drawn; the rectangle AC.BD is equal to the two rectangles AB.CD, and AD.BC. Make the angle ABE equal to the angle DBC; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC; and the angle BDA is equal E to (III. 21) the angle BCE, because they are in the same segment; therefore the triangle ABD is equiangular to the triangle BCE. Wherefore (VI. 4), BC: CE :: BD: DA, and, consequently (VÍ. 16), BC.DA=BD.CE. Again, because the angle ABE is equal to the A angle DBC, and the angle (III. 21) BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD; therefore BA: AE:: BD: DC, and BA.DC = (VI. 16) BD.AE: but it was shewn that BC.DA=BD.CE; wherefore BC.DA+BA.DC=BD.CE+BD.AE= BD.AC (II. 1). That is, the rectangle contained by BD and AC is equal to the rectangles contained by AB and ČD, and AD and BC. Therefore, the rectangles, &c. Q. E. D. PROP. E. THEOR. If an arc of a circle be bisected, and from the extremities of the arc and from the point of bisection straight lines be drawn to any point in the circumference; the sum of the two lines drawn from the extremities of the arc will have to the line drawn from the point of bisection the same ratio which the straight line subtending the arc has to the straight line subtending half the arc. Let ABD be a circle, of which AB is an arc bisected in C, and from A, C, and B to D, any point whatever in the circumference, let AD, CD, BD be drawn; the sum of the two lines AD and DB has to DC the same ratio that BA has to AC. For, since ACBD is a quadrilateral inscribed in a circle, of which the diagonals are AB and CD, AD.CB+DB.AC (VI. Ď) =AB.CD; but AD.CB+DB.AC=AD.AC+DB.AC, because CB=AC. Therefore, AD.AC+DB.AC, that B is (II. 1), (AD+DB) AC=AB.CD. And because the sides of * See Notes. equal rectangles are reciprocally proportional (VI. 14), AD+DB. DC::AB: AC. Wherefore, &c. Q. E. D. PROP. F. THEOR. If two points be taken in the diameter of a circle, such that the rectangle contained by the segments intercepted between them and the centre of the circle be equal to the square of the radius; and if from these points two straight lines be drawn to any point whatsoever in the circumference of the circle, the ratio of these lines will be the same with the ratio of the segments intercepted between the two first-mentioned points and the circumference of the circle. Let ABC be a circle, of which the centre is D, and in DA produced let the points E and F be such that the rectangle ED.DF is equal to the square of AD; from E and F to any point B in the circumference, let EB, FB be drawn; FB: BE ::FA: AE. Join BD and BA, and because the rectangle FD.DE is equal to the square of AD, that is, of DB, FD: DB::DB: DE (VI. 17). The two triangles FDB, (I. 5); but DBA is the sum of DBE and EBA, and DAB is the sum of AFB and FBA (I. 32); therefore the sum of DBE and EBA is equal to the sum of AFB and FBA; from these equals take away the equal angles DBE and AFB, and the remaining angles EBA and FBA will be equal. Thus, it appears that, in the triangle FBE, the line BA bisects the angle FBF; therefore, FB: BE:: FA: AE (VI. 3). Therefore, &c. Q. E. D. COR. The ratio of the straight lines FB, BE is also the same with the ratio of FC, CE, C being the point in which FE produced meets the circle. For, produce FB to G, and join BC. Because the angles FBE, EBG make together two right angles (I. 13), and therefore are equal to twice the sum of ABE and EBC, which make one right angle; and it has been shown that FBE is double ABE, therefore EBG is double EBC; hence it appears that the outward angle EBG is bisected by BC; therefore, FB: BE:: FC: CE (VI. A). PROP. G. THEOR. If from the extremity of the diameter of a circle a straight line be drawn in the circle, and if either within the circle, or produced without it, it meet a line perpendicular to the same diameter; the rectangle contained by the straight line drawn in the circle, and the segment of it, intercepted between the exс tremity of the diameter and the perpendicular, is equal to the rectangle contained by the diameter, and the segment of it cut off by the perpendicular. Let ABC be a circle, of which AC is the diameter, let DE be perpendicular to the diameter AC, and let AB meet DE in F; the rectangle BA.AF is equal to the rectangle CA.AD. Join BC, and because ABC is an angle in a semi B E circle, it is a right angle (III. 31). Now, the angle ADF is also a right angle (Hyp.); and the angle BAC is either the same with DAF, or vertical to it; therefore the triangles ABC, ADF are equiangular, and BA : AC:: AD: AF (VI. 4); therefore, also, the rectangle BA.AF, contained by the extremes, is equal to the rectangle AC.AD contained by the means (VI. 16). If therefore, &c. Q. E. D. The perpendiculars drawn from the three angles of any triangle to the opposite sides intersect one another in the ame point. Let ABC be a triangle, and BD, CE two perpendiculars intersecting one another in F; let AF be joined, and produced if necessary; let it meet BC in G; AG is perpendicular to BC. E F Join DE, and because AEF is a right angle, a circle described about the triangle AEF will have AF for a diameter (III. 31). Also, because ADF is a right angle, a circle described about the triangle ADF will have AF for a diameter; therefore the points A, E, F, D are in the circumference of the same B circle. And because the angles BEC, BDC are right angles, it may be shown, in the same manner, that the points B, E, D, C are in the circumference of the same circle, viz., that G |