Question

Two triangles BAC and BDC, right-angled at B and C 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.PD = CP.PB.

Answer 1

Answer:

Please find the attachment

Answer 2

Answer:

In ∆APB & ∆DPC,

∠A =  ∠D = 90°

∠APB = ∠DPC

[ vertically opposite angles]

∆APB  ~ ∆DPC

[By AA Similarity Criterion ]

AP/DP = PB/ PC

[corresponding sides of similar triangles are proportional]

AP × PC = DP × PB

Hence, proved.