Question

In the following figure, ABCD is a parallelogram then
(i) AP bisects angle A
(ii) BP bisects angle B
(iii) $\angle DAP+\angle CBP=\angle APB$

• A. True
• B. False

Answer 1

The correct option is A True

In a parallelogram,sum of adjacent angles$={180}^{\circ }$

$\angle A+\angle B={180}^{\circ }$ .......$\left(1\right)$

$AP$ bisects $\angle A$

$\angle BAP=\angle A-\angle DAP$

$BP$ bisects $\angle B$

$\angle ABP=\angle B-\angle CBP$

In $△ABP,$

$\angle BAP+\angle ABP+\angle APB={180}^{\circ }$

$\Rightarrow \angle A-\angle DAP+\angle B-\angle CBP+\angle APB={180}^{\circ }$

$\Rightarrow (\angle A+\angle B)+\angle APB={180}^{\circ }+\angle DAP+\angle CBP$

$\Rightarrow (\angle A+\angle B)+\angle APB={180}^{\circ }+\angle DAP+\angle CBP$

$\Rightarrow {180}^{\circ }+\angle APB={180}^{\circ }+\angle DAP+\angle CBP$ from $\left(1\right)$

$\therefore \angle APB=\angle DAP+\angle CBP$