Question

In the given figure, $PQRS$ is a parallelogram and $\angle SPQ={60}^{\circ }$. If the bisectors of $\angle P$ and $\angle Q$ meet at $A$ on $RS$, then which of the following is not correct?

• A. $AP=SP$
• B. $AS=AR$
• C. $AR=SP$
• D. $AQ=PQ$

Answer 1

The correct option is D $AQ=PQ$

We have $\angle SPQ={60}^{\circ }$
$\angle P+\angle Q={180}^{\circ }$ (adj. $\angle s$ of a || gm are supp)
$\angle Q={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
Now $PQ||SR$ and $AP$ is the transversal.
$\therefore SAP=\angle APQ={30}^{\circ }$ ($\because AP$ bisects $\angle SPQ$)
$\angle SPA=\angle SAP\Rightarrow SA=SP$ (isos. $\Delta$ property) ...(i)
Also $\angle RAQ=\angle AQP={60}^{\circ }$
($PQ||SR,AQ$ is transversal, alt $\angle s$)
and $\angle AQP=\frac{1}{2}\angle PQR={60}^{\circ }$
$\Rightarrow \angle RQA=\angle RAQ$ (AQ bisect $\angle PQR$)
$\Rightarrow RA=RQ$ (isos. $\Delta$ property) ....(ii)
$\therefore$ from eqn. (i) and (ii)
$AS=AR$ ($\because SP=RQ$, opp. side of a ||gm)
Also, in $\Delta ARQ,\angle ARQ={60}^{\circ }$
$\Rightarrow \Delta ARQ$ is equilateral
$\Rightarrow AR=RQ\Rightarrow AR=SP$
Therefore, the incorrect statement is $AQ=PQ$.