Question

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Which of the following should be true?

• A. $AB=BP$
• B. $\Delta APB\cong \Delta CQD$
• C. $CQ=PQ$
• D. Cannot be determined

Answer 1

The correct option is B

$\Delta APB\cong \Delta CQD$

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