ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

266 BRAHMAGUPTA AS AN ALGEBRAIST have the following solution of this problem from Brahma- gupta : The square of the diagonal (of a generated rectangle) gives three equal sides; the fourth (is obtained) by subtracting the square of the upright from thrice the square of the side (of that rectangle). If greater, it is the base; if less, it is the face.¹ As before, the rectangle generated from m, n is given by (m²–n², 2mn, m²+n²), that is these are the three sides of the right triangle, which correspond to the two sides and the diagonal of the rectangle generated by them. Let us suppose, we have a trapezium ABCD whose sides AB, BC and AD are equal, then AB = BC = AD = (m²+n²)² CD = 3(2mn)² – (m²–n²)² = 14 m²n² – m⁴ – n⁴ or CD = 3(m²–n²)² – (2mn)² = 3m⁴+3n⁴ – 10 m²n². Pṛthudaka Svāmī has taken an illustration, where m=2, n=1 and he deduces two rational trapeziums with three equal sides (25, 25, 25, 39) and (25, 25, 25, 11). The segment (CH), altitude (AH), diagonals (AC, BD) and area of this trapezium are also rational, and given by : CH (segment) = 6 m²n²—m⁴ – n⁴ AH (altitude) = 4 mn (m²—n²) AC = BD (diagonals) = 4 mn (m²+n²) ABCD (area) = 32m³n³ (m²—n²). Rational Inscribed Quadrilaterals We find in the Brāhmasphuṭasiddhānta a remarkable proposition formulated by Brahmagupta : To find all quadrilaterals which will be inscribable within circles and whose sides, diagonals, perpendi- culars, segments (of sides and diagonals by perpendi- culars from vertices as also of diagonals by their intersection), areas, and also the diameters of the

1. कर्णकृतित्रिसम भुजास्त्रयश्चतुर्थो विशोष्य कोटि कृतिम् । बाहुकृतेस्त्रिगुणाया यद्यधिको भूर्मुखं हीनः ॥ —BrSpSi. XII. 37