Question

In trapezium $ABCD$, $AD\parallel BC$. Diagonal $AC$ and diagonal $BD$ intersect each other in point $P$. Then show that $\frac{AP}{PD}=\frac{PC}{BP}$.

Answer 1

In $△PAD$ and $△PCB$,

$\angle APD=\angle BPC$ [Vertically opposite angle]

$\angle PAD=\angle PCB$ [Alternate angles]

Hence by $AA$ criterion of similarity,

$△PAD\sim △PCB$

So, by basic proportionality theorem,

$\frac{AP}{PC}=\frac{PD}{PB}$

$\Rightarrow \frac{AP}{PD}=\frac{PC}{BP}$

Hence proved.