Question

Two right triangles abc and dbc are drawn on the same hypotenuse bc and on the same side of bc. If ac and bd intersect at p, prove that ap x pc bp x dp.

Answer 1

Please see 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 = PB × DP

Hence, proved.