Mathematics in Ancient and Medieval India c6.s9.60

c6.s9.60

= a² + b² - 2ab.(a² + b² - c² - d²)/(2(ab + cd)) [ From 4 ] = (2(ab + cd)(a² + b²) - 2ab(a² + b² - c² - d²))/(2(ab + cd)) = (2cd(a² + b²) + 2ab (c² + d²))/(2(ab + cd)) = ((ac + bd)(ad + bc))/(ab + cd) ∴ AC = d₁ = √(((ac + bd) (ad + bc))/(ab + cd)) Similarly it may be shown, BD = d₂ = √(((ac + bd)(ab + cd))/(ad + bc)) Again, d₁².d₂² = (ac + bd)² ∴ d₁.d₂ = ac + bd.