Question

$ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal.
$BD($ see Fig.$).$ Show that $AP=CQ$.

Answer 1

From the given figure

$\begin{array}{c}In\Delta APBand\Delta CQD\\ \angle APB=\angle CQD={90}^{\circ }\\ AB=CD\\ and,\\ \angle ABP=\angle CDQ\left(alternateanglesareequal\right)\\ Hence,the\Delta APB\simeq \Delta CQD\\ Hence,\\ AP=CQ.\end{array}$