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 (5. 4.) the circle ACB about the triangle, and draw its diameter AE, and join EC : Because the right angle BDA is equal (3. 3.) to the angle ECA in a semicircle, and the angle ABD to B C the angle AEC, in the same segment D (21. 3.) ; the triangles ABD, AEC are equiangular : Therefore, as (4. 6.) BA to AD, so is EA to AC: and consequently the rectangle BA.AC is equal (16. 6.) to the rectangle EA.AD. If, E therefore, from an angle, &c. Q. E. 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 to (21. 3.) the angle BCE, because they are in the same segment ; therefore the triangle ABD is equian- B gular to the triangle BCE. Wherefore (4. 6.), BC : CE :: BD : DA, and consequently (16. 6.) BC.DA = BD.CE. Again, because the angle ABE is equal to the angle DBC, and the angle (21. 3.) BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD; therefore BA : AE :: BD : DC, and A D BA.DC = BD.AE: But it was shewn that BC.DA =BD.CE; wherefore BC. DA + BA.DC = BD.CE + BD.AE = BD.AC (1. 2.). That is the rectangle contained by BD and AC, is equal to the rectangles contained by AB, CD, and AD, BC. Therefore the rectangle, &c. Q. E. D. X ᅮ PROP. E. THEOR. If an arch of a circle be bisected, and from the extremities of the arch, 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 arch will have to the line drawn from the point of bisection, the same ratio which the straight line subtending the arch has to the straight line subtending half the arch. Let ABD be a circle, of which AB is an arch 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. D For since ACBD is a quadrilateral inscribed in a circle, of which the diagonals are AB and CD, AD.CB + DB.AC (D. 6.) = AB.CD : but AD. CB+DB.AC=AD.AC+DB.AC, be- . cause CB=AC. Therefore AD.AC+DB.AC, that is (1.2.), (AD + DB) AC=AB.CD. And because the sides of equal rectangles are reciprocally A B proportional (14. 6.), 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 I 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 because the rectangle FD, DE is equal to the square of AD, that is, of DB, FD : DB :: DB : DE (17. 6.). The two triangles, FDB, BDE have therefore the sides proportional that are about the common angle D ; therefore they are equiangular (6. 6.), the angle DEB being equal to the angle DBF, and DBE to DF). Now since the sides about these equal angles ate also propor ional (4. 6.), FB : BD : : BE : ED, and alternately (16. 5.), FB :: BE :: BD : ED, or FB : BE :: AD: DE. But because FD: DA : : DA : DE, by division (17. 5.), FA : DA :: AE : ED, and alternately (11. 5.), FĂ : AE :: DA : ED. Now it has been shewn that FB : BE:: AD : DE, therefore FB : BE :: FA : AE. Therefore, &c. Q. E. D. Cor. If AB be drawn, because FB : BE :: FA : AE, the angle FBE is bisected (3.6.) by AB. Also, since FD : DC :: DC : DE, by composition (18. 5.), FC : DC :: CE : ED, and since it has been shewn that FÀ : AD (DC): : AE : ED, therefore, ex æquo, FA : AE: FC :: CE. But FB : BE :: FA : AE, therefore, FB : BE :: FC: CE (11.5.); so that if FB be produced to G, and if BC be drawn, the angle EBG is bisected by the line BC (A 6.). 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 with- E B B A F A ID pendicular to the diameter AC, and let AB meet DE in F; the rect-angle BA.AF is equal to the rectangle CA.AD. Join BC, and because ABC is an angle in a semicircle, it is a right angle (31. 3.): 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 (4.6.); therefore also the rectangle BĂ.AF, contained by the extremes, is equal to the rectangle AC.AD contained by the means (16. 6.). If therefore, &c. Q E. D. PROP. H. THEOR. The perpendiculars drawn from the three angles of any triangle to the opposite sides intersect one another in the same point. Let ABC be a triangle, BD and 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. Join DE, and about the triangle AEF let a circle be described, AEF ; then, because AEF is a right angle, the circle described about the triangle AEF will have AF for its diameter (31. 3.). In the same manner, the circle described about the triangle ADF has AF for its diameter ; therefore the points A, E, F and D are in the circumference of the same circle. But because the angle EFB is equal to the angle DFC E (15. 1.), and also the angle BEF E to the angle CDF, being both right angles, the triangles BEF and CDF are equiangular, and therefore BF : EF :: CF: FD (4. 6.), B G or alternately (16. 5.) BÈ : FC :: EF : FD. Since, then, the sides about the equal angles BFC, EFD are proportionals, the triangles BFC, EFD are also equiangular (6. 6.); wherefore the angle FCB is equal to the angle EDF. But, ÈDF is equal to EAF, because they are angles in the same segment (21. 3.) ; therefore the angle EAF is equal to the angle FCG: Now, the angles AFE, CFG are also equal, because they are vertical angles; therefore the remaining angles AEF,FGC are also equal (32. 1.): But AEF is a right angle, therefore FGC is a right angle, and 4G is perpendicular to BC. Q. E. D. Cor. The triangle ADE is similar to the triangle ABC. For the two triangles BAD, CAE having the angles at D and E right angles, and the angle at A common, are equiangular, and therefore BA : AD : CA : AE, and alternately BA : CA :: AD : AE ; therefore the two, triangles BAC, DAE, have the angle at A common, and the sides about С that angle proportionals, therefore they are equiangular (6. 6.) and similar. Hence the rectangles BA.AE, CA.AD are equal. If from any angle of a triangle a perpendicular be drawn to the opposite side or base ; the rectangle contained by the sum and difference of the other two sides, is equal to the rectangle contained by the sum and difference of the segments, into which the base is divided by the perpendicular. Let ABC be a triangle, AD a perpendicular drawn from the angle A on the base BC, so that BD, DC are the segments of the base ; (AC+AB) (AC-AB)=(CD+DB) (CD-DB). From A as a centre with the radius AC, the greater of the two sides, describe the circle CFG ; produce AB to meet the circumference in E and F, and CB to meet it in G. Then because AF=AC, BF=AB+AC, the sum of the sides ; and since AE=AC, BE=AC -AB-= the difference of the sides. Also, because AD drawn from the centre cuts GC at right angles, it bisects it ; therefore, when the perpendicular falls within the triangle, BG=DG - DB=DC—DB= the difference of the segments of the base, and BC=BD+DC=the sum of the segments. But when AD falls without the triangle, BG= DG+DB=CĎ+-DB=the sum of the segments of the base, and BC =CD-DB = the difference of the segments of the base. Now, in both cases, because B is the intersection of the two lines FE, GC, drawn in the circle, FB.BE=CB.BG ; that is, as has been shown, (AC+AB) (AC--AB) = (CD+DB) (CD-DB). Therefore, &c. |