Mathematics in Ancient and Medieval India c6.s10.8

c6.s10.8

¹ Brahmagupta (*Brāhmasphuṭasiddhānta*, ch. 12, v. 34) divides the scalene triangle ABC (Fig. 15) by drawing a perpendicular AD from the vertex A. The perpendicular AD evidently divides the triangle ABC into two right-triangles ADB and ADC. If AD=p, then from the previous knowledge, AB = ½((p²)/m+m), BD=½((p²)/m-m) and AC=½((p²)/n+n) and DC=½((p²)/n-n) where m, n are rational numbers. Hence BD = ½((p²)/m+(p²)/n-m-n). Mahāvīra (850 A.D.) says (Datta, B. and Singh, A.N. *Ibid.*, 2, p. 225-227) that if m²-n², 2mn, m²+n² and p²-q², 2pq, p²+q² be the sides of the two rightangled triangle and AD is so chosen that AD = 2mn=2pq, then the two triangles may be taken as ABD, ACD or as AED, ACD where ABD=AED. Then the triangle is ABC or AEC. The problem and its solutions are given in Europe by Buchet (1621 A.D.) and Cunliffe (Dickson, *History of Theory of Numbers*, 2, p. 192-3). the opposite sides of another convex quadrilateral (Vide *Brāhmasphuṭasiddhānta*, ch. 12, v. 38). Brahmagupta previously mentioned the term *caturbhūjakoṇaspṛsagvṛtta* which means that he probably considered "an inscribed quadrilateral whose vertices touch the circle" (*Ibid.*, ch. 12, v. 27).