ABCD is a parallelogram AP and CQ are perpendicular drawn from vertices A and C on diagonal BD (see figure) show that AP = CQ.

Hint: We will first start by using the fact that if two triangles are congruent then their corresponding parts are equal for proving two triangles congruent. We will use AAS congruency wherein we have to prove two angles and one side are equal.

Complete step-by-step solution -
Now, we have been given a parallelogram and AP and CQ are perpendicular drawn from vertices A and C and diagonal BD and we have to show that AP = CQ.

Now, we know that opposite sides of a parallelogram are parallel and equal. Therefore, we have,
$AD=BC ......................\left( 1 \right)$
Also, we have $AD\parallel BC$ and we know that alternate interior angles formed by a transversal cutting a pair of parallel lines are equal. Therefore we have,
$\angle 1=\angle 2 ...............\left( 2 \right)$
Now, we know that the angle formed by a straight line is $180{}^\circ $. Therefore, we have,
$\begin{align}
  & \angle DPA+90{}^\circ =180{}^\circ \\
 & \Rightarrow \angle DPA=90{}^\circ \\
 & \angle CQB+90{}^\circ =180{}^\circ \\
 & \Rightarrow \angle CQB=90{}^\circ \\
 & \Rightarrow \angle DPA=\angle CQB .................\left( 3 \right) \\
\end{align}$
Now, we have in $\Delta DPA$ and $\Delta BQC$,
$\begin{align}
  & \angle DPA=\angle BQC \\
 & AD=CB \\
 & \angle 1=\angle 2 \\
\end{align}$
Hence, $\Delta DPA\cong \Delta BQC$ by AAS congruency.
Now, we know that by the corresponding part of the congruent triangle we have,
AP = CQ
Hence proved.

Note: It is important to note that we have used a fact that the sum of the linear pair of angles is $180{}^\circ $. Therefore, we have $\angle APD=\angle CQB=90{}^\circ $. Also, we have used a fact that the opposite sides of a parallelogram are equal and parallel. It is also worthwhile to remember that the alternate interior angle formed by a transversal on a pair of parallel lines are equal.