Question

In ∆ABC, AD is perpendicular to BC and ADsquare = BD×CD then prove that ∆ABC is a right angled triangle

Answer 1

I hope it will help you

Given
InΔABC,AD⊥BCandIn∆ABC,AD⊥BCandAD²=BD.DC

Now,

InΔABD,AB²=AD²+BD²

InΔADC,AC²=AD²+CD²

(1)+(2)
⟹AB²+AC²=2AD²+BD²+CD²

⟹AB²+AC²=2BD.DC+BD²+CD²[Given,AD²=BD.DC

⟹AB²+AC²=(BD+CD)²

⟹AB²+AC²=BC²


So that, according to the PYTHAGORAS law ∠BAC=90°.

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Answer 2

Answer:

Check your answer please