Question

two right angled triangle abc and dbc are on the same hypotenuse bc side ac and bd meet at p prove that ap *pc=pd*bp

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 = DP × PB

Hence, proved.