History of Hindu Mathematics: A Source Book v2.c17.s19.6

v2.c17.s19.6

AB &= 1/2((4m²n²)/p-p) - (m²-n²), \\ AD &= BC = m²+n²; \end{aligned} also DH &= m²-n², / HC &= 1/2((4m²n²)/p-p), / AC &= BD = 1/2((4m²n²)/p+p), / AH &= 2mn, area ABCD = mn((4m²n²)/p-p).