Question

Right angles triangles BAC and BDC are right angled at A and D and they are on the same side of BC. If AC and BD intersect at P, then prove that AP.PC = PB.DP. Can someone explain please?

Answer 1

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.