Question

Given : In the fig < ACB = 90°. and CD is parallel to AB
to prove: CB²/ CA² = BD/AD​

Answer 1

Answer:

Step-by-step explanation:

On 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

Step-by-step explanation:

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Answer 2

the upper attachment really help u...