Question

ABCDand ABCD are right-angled triangles. _BAC = 90°. _BDC = 90°, and AC = BD. Prove that ∆ABC~ ∆ABC
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Answer 1

Explanation:

ANSWER

In △APB and △DPC, we have

 ∠A=∠D=90∘     and

 ∠APB=∠DPC        [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

    △APB∼△DPC

⇒ DPAP=PCPB

⇒  AP×PC=DP×PB              [Hence proved]