Question

Quadrilateral $ABCD$ is a square. $P$, $Q$ and $R$ are the points on $AB$, $BC$ and $CD$ respectively, such that $AP=BQ=CR$. Hence,.If $\angle$ $PQR$ is a right angle, find $\angle PRQ$(in degrees)

Answer 1

Given: ABCD is a square.
thus, $AB=BC$
$AP=BQ$ (Given)
$AB-AP=BC-BQ$
$PB=CQ$ (I)
Now, In $△PBQ$ and $△CQR$,
$\angle PBQ=\angle QCR$ (Each ${90}^{\circ }$)
$PB=CQ$ (From I)
$BP=CR$ (Given)
thus, $△PBQ\cong △CQR$ (SAS rule)
or, $PQ=QR$ (By cpct)
Now, In $△PQR$
$PQ=QR$
thus, $\angle PRQ=\angle QPR=x$ (Isosceles triangle property)
$\angle PQR={90}^{\circ }$ (Given)
Sum of angles = 180
$\angle PQR+\angle QPR+\angle QRP=180$
$90+x+x=180$
$2x=90$
$x=45$
Thus, $\angle PRQ={45}^{\circ }$