Question

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD( See the given figure). Show that
$\Delta APB\cong \Delta CQD$

Answer 1

In $\Delta APB\cong \Delta CQD$
$\angle APB=\angle CQD(Each{90}^{\circ }$)
AB=CD (Opposite sides of parallelogram ABCD)
$\angle ABP=\angle CDQ$ (Alternate interior angles for AB || CD)
$\therefore \Delta APB\cong \Delta CQD$ (By AAS congruency)