Question

Bisectors of angles A and D of a quadrilateral ABCD meet at P. then,
$\angle APD=\frac{1}{2}[\angle B+\angle C].$

State true or false.

• A. True
• B. False

Answer 1

The correct option is A True

In $△APD$
$\angle APD+\angle PAD+\angle PDA=180$
$\angle APD=180-\frac{1}{2}(\angle A+\angle D)$
Now, in quadrilateral ABCD,
$\angle A+\angle B+\angle C+\angle D=360$
$\angle B+\angle C=360-\angle A-\angle D$
$\frac{1}{2}(\angle B+\angle C)=180-\frac{1}{2}(\angle A+\angle D)$
$\frac{1}{2}(\angle B+\angle C)=\angle APD$