Question

$ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ Show that
i) $△APB\cong △CQD$
ii) $AP=CQ$

Answer 1

(i) Since $ABCD$ is a parallelogram. Therefore, $DC\parallel AB$.

Now $DC\parallel AB$ and transversal $BD$ intersects them at $B$ and $D$.

Therefore, $\angle ABD=\angle BDC$ [ Alternate interior angles]

In $△APB$ and $△CQD$, we have

$\angle ABP=\angle QDC$ ....(alternate interior angles of parallelogram $ABCD$ and $DC\parallel AB$)

$\angle APB=\angle CQD$ ....[each angle ${90}^{\circ }$]

and $AB=CD$ [Opposite. sides of a || gm]

Therefore, by AAS criterion of congruence $△APB\cong △CQD$

(ii) Since $△APB\cong △CQD$

Therefore, $AP=CQ$ $\mid$ Since corresponding parts of congruent triangles are equal