Question

In parallelogram $ABCD$, the bisector of angle $A$ meets $DC$ at $P$ and $AB=2AD$.
Prove that :
i) $BP$ bisects angle $B$.
ii) Angle $APB={90}^o$.

Answer 1

(i) Let $AD=x,AB=2,AD=2x$

Also AP is the bisector $\angle A\therefore \angle 1=\angle 2$

Now, $\angle 2=\angle 5$ (alternate angles)

$\therefore \angle 1=\angle 5NowAD=DP=x$ [$\because$ Sides opposite to equal angles are also equal]

$\because AB=CD$ (opposite sides of parallelogram are equal)

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

Also,$BC=x$ in $\Delta BPC,\angle 6=\angle 4$ (Angles opposite to equal sides are equal)

Also, $\angle 6=\angle 3$ (alternate angles)

$\because \angle 6=\angle 4$ and $\angle 6=\angle 3\Rightarrow \angle 3=\angle 4$

Hence, BP bisects $\angle B$.

(ii) To prove $\angle APB=$ 90°$\because$ Opposite angles are supplementary..

Angle sum property.