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arcb BL be equal to EN, the sector BGL is equal to the sector EHN;
and if the arch 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 arches BC, EF, and the two sectors BGC, EHF, and of the arch BC, and sector BGC, the arch BL and the sector BGL are any equimultiples whatever; and of the arch EF, and sector EHF, the arch EN and sector EHN, are any equimultiples whatever; and it has been proved, that is the arch BL be greater than EN, the sector BGL is greater than the sector EHN; if equal, equal ; and if less, less; therefore (def. 5.5.), as the arch BC is to the arch EF, so is the sector BGC to the sector EHF. Wherefore, in equat circles, &e. Q. E. D.
PROP. B. THEOR. 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.
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 (5. 4.) 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 (21. 3.) AEC, for they are in the same segment; the triangles ABD, AEC are equiangular to one another: Therefore BA : AD :: EA : (4. 6.) AC, and consequently, BA.AC (16. 6.) AD.AE = ÈD.DA (3. 2.) +DA. But ED.DA= BD.DC, therefore BA.AC = BD.DC + DA%. Wherefore, if an angle, c. Q. E. D.
PROP. C. THEOR. 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 BDĂ is equal (3. 3.) to the angle ECA in a semicircle, and the angle ABD to the angle AEC, in the same segment (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, therefore, from an angle, &c. Q. E. D.
PROP. D. THEOR. 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 cirele, 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.) BCDA = 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 BA.DC=BD.AE: But it was shown that BC.DA=BD.CE; wherefore BC. DA+BA.DC=BD.CE+BD.AE=BD.AC (1. 2.). That is the l'ectangle contained by BN and AC, is equal to the rectangles cons
tained by AB, CD, and AD, BC. Therefore the rectangle, &c. Q. E. D.
PROP. E. THEOR. If an arch of a circle be bisected, and from the ex. tremities 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 cireumference, 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 A CBD 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, because CB=AC. Therefore ADAC + DB.AC, that is (1. 2.), (AD+DB) AC=AB.CD. And because the sides of equal rectangles are reciprocally 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 Kines be drawn to any point whatsoever in the circumference of the eircle, 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 ce tre is D, and in DA produced, let the points Eand F be such that th: 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 10 DFB. Now since the sides about these equal angles are also propor
tional (4. 6.), FB : BD :: BE : ED, and alternately (16, 5.), FB : BE :: BD: ED, or FB : BE:: AD : DE. But because FD: ÞA:: DA: DE, by division (17.5.), FA:DA :: AE: ED, and alternately (11.5.), FA : AE :: DA: ED. Now it has been shown 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 shown that FA:AD (DC) :: AE:ED, therefore, ex æquo, FA : AE: FC :: CE. But FB : BÉ:: 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 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 extremity 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 a 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 be cause ABC is an angle in a semicircle, it is a right angle (31. 3.) : Now, the angle ADĚ 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 BA.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 he 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, hecause AEF is a ri_ht angle, the circle described about the triangle AEF will have AF for its diameter (31. 3.). In the same manner, the circle describ. ed 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 (15. 1.), and also the angle BEF to the angle CDF, being both right angles, the triangles BEF and CDF are equiangular, and therefore BF:EF:, CF ;FD (4. 6.), B or alternately (16. 5.) BF : 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 EDF 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 an: