Question

In Fig. 6.48, ∠ACB = 90° and CD⟂AB. Prove that BC^2/AC^2=BD/AD

Answer 1

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QUESTION :)




In Fig. 6.48, ∠ACB = 90° and CD⟂AB. Prove that BC^2/AC^2=BD/AD



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ANSWERS :)





ΔACD ~ΔABC (Theorem 6.7)





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

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#Be brainly

Answer 2

Answer:

See the attachment. Hope it helps uh!