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 BAC be bi sected by AD; then BA × AC=BD × DC + AD'. Describe (E. v. 4.) the circle ACB about the triangle; produce AD to the circumference in E, and draw EC. And, because (E. xxi. 3.) the ABC = ▲ AEC, and (hyp.) the ▲ BAD= ¿CAE; therefore (E. xxxii. 1.) the as ABD, AEC, are equiangular; therefore (E. iv. 6.) BA: AD :: EA: AC; BA × AC=EA × AD; that is, (E. iii. 2.) BA × AC=ED × DA + AD3: But (E. xxxv. 3.) ED × DA=BD × DC; therefore BA × AC = BD × DC + AD2. PROP. LVI. 65. THEOREM. 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 perpendi cular and the diameter of the circle described about the triangle. Let ABC be a triangle, and AD the perpendicular B from the BAC to the base BC; then is BA × AC equal to the rectangle contained by AD, and the diameter of the circle described about the ▲ ABC. Describe (E. v. 4.) the circle ACB about the triangle; draw its diameter AE, and join E, C: Because the ECA in a semi-circle is equal (E. xxxi. 3.) to the right BDA, and that (E. xxi. 3.) the AEC=▲ ABC; therefore (E. xxxii. 1.) the As ABD, AEC are equiangular, therefore (E. iv. 6.) BA: AD :: EA: AC; PROP. LVII. 66. THEOREM. The rectangle contained by the diagonals of a quadrilateral rectilineal figure, inscribed in a circle, is equal to both the rectangles contained by its opposite sides. This is Prop. D. Book VI. of Euclid's Elements, as edited by Simpson. PROP. LVIII. 67. THEOREM. If, from the centre of the circle, described about a given triangle, perpendiculars be drawn to the three sides, their aggregate shall be equal to the radius of the circumscribed circle, together with the radius of the circle inscribed in the given triangle. Let ABC be the given triangle; bisect (E. x. 1.) AB, BC, and AC in the points D, E, and F; and from D, E, and F draw (E. xi. 1.) DG perpendicular to AB, EG perpendicular to BC, and FG perpendicular to AC; then (Supp. iv. 1.) these perpendiculars meet in the same point G, which is the centre of the circle that can be described about the ▲ ABC; find, also, (E. iv. 4.) the centre K, and the semi-diameter KH, of the circle that can be inscribed in the ▲ ABC; and draw GA: Then* GD+GE+GF=GA+KH. * The straight lines GD, GE, GF, are to be supplied in the figure. For draw DE, EF, and FD; therefore (Supp. lxix. 1. Cor. 1. and E. xxxiv. 1.) 1⁄2 AC, CF=AB, and FD= BC; draw GB, and GC; And, since (constr.) the angles at D, E, F, are right angles; therefore (E. xxxii. 1. Cor. 1.) the two s DAF, DGF, are, together, equal to two right angles; therefore (Supp. xxviii. 3.) a circle may be described about the trapezium ADGF; and in the same manner it may be shewn that circles may be described about BDGE, and CFGE: therefore (Supp. lvii. 6.) AG × DF+BG × DE+ CG FE = AF × DG + AD × GF+BD × GE+ BE × DG + CEx GF + CF × GE: And if to the doubles of these equals be added the rectangles GE × BC+GF × AC+GD × AB, which (E. xli. 1.) make up the double of the ▲ ABC, it will be manifest, from E. i. 2., that the rectangle contained by the perimeter of the ▲ ABC, and by GA, together with the double of the ▲ ABC, is equal to the rectangle contained by the perimeter of ABC, and by the aggregate of GD, GE, and GF: But (Supp. ii. 4.) the double of the ▲ ABC is equal to the rectangle contained by the perimeter of the triangle and the semi-diameter, KH, of the circle inscribed in it; therefore (E. i. 2.) the rectangle contained by the perimeter, and by the aggregate of GA and KH, is equal the rectangle contained by the perimeter, and by the aggregate of GD, GE, and GF; therefore GD+GE+GF=GA+KH. PROP. LIX. 68. PROBLEM. To find a point, from which if three straight lines be drawn to three given points, their differences shall be severally equal to three given straight lines; the difference of any two of the straight lines to be drawn, not being greater than the distance of the two points to which they are to be drawn. Let A, B, C, be the three given points, and R, S, two of the given differences: It is required to find a point, from which if three straight lines be drawn to A, B, and C, the difference of the first and second shall be equal to R, the difference between the second and third equal to S, and therefore the difference between the first and third equal to the third of the given differences. Draw AB, BC, and CA; bisect (E. x. 1.) AB in D, and BC in E; from DB cut off DF, equal to a third proportional (E. xi. 6.) to 2 AB, and to S; likewise from EB cut off EG, equal to a third proportional to 2 BC, and to R; and through F and G draw (E. xi. 1.) FH perpendicular to AB, and GH |