$ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ (see Fig). Show that $\Delta APB \cong \Delta CQD$.
(N/A) In $\Delta APB$ and $\Delta CQD$,we have:
$1$. $\angle APB = \angle CQD = 90^{\circ}$ (Given,as $AP \perp BD$ and $CQ \perp BD$)
$2$. $AB = CD$ (Opposite sides of a parallelogram are equal)
$3$. $\angle ABP = \angle CDQ$ (Alternate interior angles,since $AB \parallel CD$ and $BD$ is a transversal)
Therefore,by the $AAS$ (Angle-Angle-Side) congruence criterion,we have:
$\Delta APB \cong \Delta CQD$
$1$. $\angle APB = \angle CQD = 90^{\circ}$ (Given,as $AP \perp BD$ and $CQ \perp BD$)
$2$. $AB = CD$ (Opposite sides of a parallelogram are equal)
$3$. $\angle ABP = \angle CDQ$ (Alternate interior angles,since $AB \parallel CD$ and $BD$ is a transversal)
Therefore,by the $AAS$ (Angle-Angle-Side) congruence criterion,we have:
$\Delta APB \cong \Delta CQD$
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