Question

In the figure,seg $AC$ and seg $BD$ intersect each other in point $P$ such that $\frac{AP}{CP}=\frac{BP}{DP}$.Prove that, $△ABP\sim △CDP$

Answer 1

IN figure it is given that seg$AC$ and seg$BD$ intersect each other in point $P$ such that $\frac{AP}{CP}=\frac{BP}{DP}$

$In\Delta APBand\Delta DCP\frac{AP}{CP}=\frac{BP}{DP}\left(given\right)\angle APB=\angle DPC\left(verticallyoppositeangle\right)\therefore \Delta ABP\sim \Delta CDP\left(SAS\right)$