Question

$ABCD$ is a parallelogram. $AP$ and $CQ$ are perpendicular lines drawn from vertices $A$ and $C$ respectively on diagonal $BD$ (see figure). Show that
$△APB\cong △CQD$

Answer 1

Given : $ABCD$ is a parallelogram. $AP$ and $CQ$ are perpendicular lines drawn from vertices A and C respectively on diagonal $BD$.

To show : $\Delta APB\cong \Delta CQD$

Proof:

In triangle $APB$ and triangle $CQD$.

$AB=CD$ (Opposites sides of parallelogram)

$\angle ABP=\angle CDQ$ ( Alternate interior angles)

$\angle APB=\angle CQD={90}^{\circ }$ (Given )

Therefore, by $AAS$ criterion of congruency, triangle $APB$ is congruent to triangle $CQD$.