Question

Two triangles BAC and BDC right angled at A and D respectively are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P. Prove that AP PC=DP X PB.​

Answer 1

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Explanation:

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Answer 2

Answer:

Solution :

In △APB and △DPC, we have

∠A=∠D=90 degree

and

∠APB=∠DPC [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

△APB∼△DPC

⇒

DP

AP

=

PC

PB

⇒ AP×PC=DP×PB [Hence proved]