Question

State true or false:

In parallelogram $ABCD$, the bisector of angle $A$ meets $DC$ in $P$ and $AB=2AD$, then

$\angle APB={90}^0$

• A. True
• B. False

Answer 1

The correct option is A True

We have to prove that $\angle APB=90°$

Opposite angles are supplementary

$\Rightarrow \angle 1+\angle 2+\angle 3+\angle 4=180°$ $[\angle 1=\angle 2,\angle 3=\angle 4]$

$\Rightarrow 2\angle 2+2\angle 3=180°$

In $△APB$

$\Rightarrow \angle 2+\angle 3+\angle APB=180°$

$\Rightarrow \angle APB=180°-90°=90°.$

Hence, the statement is true.