Question

in triangle ABC, AD perpendicular to BC,such that AD2=BD.CD prove that triagle ABC is right triangle at A

Answer 1

Prove:Given,Prove:Given,

InΔABC,AD⊥BCandIn∆ABC,AD⊥BCandAD²=BD.DCAD²=BD.DC

Now,Now,

InΔABD,AB²=AD²+BD²→.1In∆ABD,AB²=AD²+BD²→.1

InΔADC,AC²=AD²+CD²→.2In∆ADC,AC²=AD²+CD²→.2

(1)+(2)⟹AB²+AC²=2AD²+BD²+CD²(1)+(2)⟹AB²+AC²=2AD²+BD²+CD²

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

⟹AB²+AC²=(BD+CD)²⟹AB²+AC²=(BD+CD)²

⟹AB²+AC²=BC²⟹AB²+AC²=BC²