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 × PB.

Answer 1

In $△APBand△DPC$, we have

$\angle APB=\angle DPC$

$\angle BAC=\angle BDC$

so $△APB=△DPC$

Corresponding sides will be proportional

so $AP\times PC=DP\times PB$