1. If a line CD bisect the vertical angle C of any triangle ACB, its square added to the rectangle AD.DB contained by the segments of the base is equal to the rectangle contained by the sides. Dem.—Describe a circle about the triangle, and produce CD to meet it in E; then it is easy to see that the triangles ACD, ECB are equiangular. Hence [iv.] AC : CD :: CE : CB; therefore AC.CB = CE.CD = CD2 + CD.DE = CD2 + AD.DB [III. xxxv.]. 2. If the line CD′ bisect the external vertical angle of any triangle ACB, its square subtracted from the rectangle AD′.D′B is equal to AC.CB. 3. The rectangle contained by the diameter of the circumscribed circle, and the radius of the inscribed circle of any triangle, is equal to the rectangle contained by the segments of any chord of the circumscribed circle passing through the centre of the inscribed. Dem.—Let O be the centre of the inscribed circle. Join OB (see foregoing fig.); let fall the perpendicular OG, draw the diameter EF of the circumscribed circle. Now the angle ABE = ECB [III. xxvii.], and ABO = OBC; therefore EBO = sum of OCB, OBC = EOB. Hence EB = EO. Again, the triangles EBF, OGC are equiangular, because EFB, ECB are equal, and EBF, OGC are each right. Therefore, EF : EB :: OC : OG; therefore EF.OG = EB.OC = EO.OC. 4. Ex. 3 may be extended to each of the escribed circles of the triangle ACB. 5. The rectangle contained by two sides of a triangle is equal to the rectangle contained by the perpendicular and the diameter of the circumscribed circle. For, let CE be the diameter. Join AE. Then the triangles ACE, DCB are equiangular; hence AC : CE :: CD : CB; therefore AC.CB = CD.CE. 6. If a circle passing through one of the angles A of a parallelogram ABCD intersect the two sides AB, AD again in the points E, G and the diagonal AC again in F; then AB.AE + AD.AG = AC.AF. Dem.—Join EF, FG, and make the angle ABH = AFE. Then the triangles ABH, AFE are equiangular. Therefore AB : AH :: AF : AE. Hence AB.AE = AF.AH. Again, it is easy to see that the triangles BCH, FAG are equiangular; therefore BC : CH :: AF : AG; hence BC.AG = AF.CH, or AD.AG = AF.CH; but we have proved AB.AE = AF.AH. Hence AD.AG + AB.AE = AF.AC. 7. If DE, DF be parallels to the sides of a triangle ABC from any point D in the base, then AB.AE + AC.AF = AD2 + BD.DC. This is an easy deduction from 6. 8. If through a point O within a triangle ABC parallels EF, GH, IK to the sides be drawn, the sum of the rectangles of their segments is equal to the rectangle contained by the segments of any chord of the circumscribing circle passing through O. Dem.—AO.AL = AB.AK + AC.AE. (6) But AO2 = AG.AK + AH.AE − GO.OH. (7) Hence AO.OL = BG.AK + CH.AE + GO.OH, or AO.OL = EO.OF + IO.OK + GO.OH.