Question

In fig angle ACB= 90* and CD perpendicular to AB. Prove that
$\frac{ {bc}^{2} }{{ac} ^{2}} = \frac{ {bd} }{ ad}$

Answer 1

In triangleABC and triangleACD

∠ACB = ∠CDA,

∠CDA = CAB

ΔABC and ΔACD

 

AC/AB=AD/AC

AC²= AB×AD

similarly ΔBCD and ΔBAC

BC/BA = BD/BC

BC²= BA×BD

∴ ÷ BC² and AC² we get

BC²/AC²=AB×BD/AB×AD

∴ BC²/AC² =BD/AD

hence proved

Answer 2

Hope it helps you....