Question

State true or false:

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

$BP$ bisects angle $B$.

• A. True
• B. False

Answer 1

The correct option is A True

Let $AD=x$ $;$ $AB=2AD=2x$

Also $AP$ is the bisector $\angle A$

$\therefore \angle 1=\angle 2$

$\therefore \angle 2=\angle 5$ [alternate angles ]

$\Rightarrow \angle 1=\angle 5$

$AD=DP=2$

[ Sides opposite to equal angles are also equal ]

$\therefore AB=CD$ ( Opposite sides of parallelogram are equal )

$\therefore CD=2x$

$\Rightarrow DP+PC=2x$ $x+PC=2x$

$\therefore PC=x$

Also, $BC=x$

In $△BPC,$ $\angle 6=\angle 4$ [ Angles opposite to equal sides are equal ]

$\Rightarrow \angle 6=\angle 3$ [alternate anlges ]

$\therefore \angle 6=\angle 4$ and $\angle 6=\angle 3$

$\Rightarrow \angle 3=\angle 4$

Hence $BP$ bisects $\angle B.$

Hence, the answer is the statement is true.