Question

In the given figure, ABCD is a square and $\angle PQR={90}^{\circ }$. If PB = QC = DR, prove that (i) QB = RC, (ii) PQ = QR, (iii) $\angle QPR={45}^{\circ }$

Answer 1

(i)

ABCD is a square

$BC=CD$

$GivenQC=RD$

$\therefore BC-QC=CD-RD$

$\Rightarrow BQ=CR$

(ii)

In $\Delta PBQ$ and \\Delta QCR\)

$PB=QC$ [given]

$BQ=CR$ [Proved in (a)]

$\angle PBQ=\angle QCR$ [Each is ${90}^{\circ }$]

$\therefore \Delta PBQ$ and $\Delta QCR$ [SAS Congruency]

$\Rightarrow PQ=QR,\angle BPQ=\angle CQR,\angle BQP=\angle CRQ$ [C.P.C.T.]

BQC is a straight line

$\therefore \angle BQP+\angle PQR+\angle CQR={180}^{\circ }$

$\Rightarrow \angle BQP+\angle PQR+\angle BPQ={180}^{\circ }$

$\Rightarrow (\angle BQP+\angle BPQ)+\angle PQR={180}^{\circ }$

$\Rightarrow {180}^{\circ }-\angle PBQ+\angle PQR={180}^{\circ }$ [Angle sum property for $\Delta PBQ$]

$\Rightarrow {180}^{\circ }-{90}^{\circ }+\angle PQR={180}^{\circ }$

$\Rightarrow \angle PQR={90}^{\circ }$

(iii)

$\Delta PQR$ is an isosceles right angled triangle, as $PQ=QR$ and $\angle PQR={90}^{\circ }$.

$\Rightarrow \angle PQR+\angle QRP+\angle QPR={180}^{\circ }$

$\Rightarrow {90}^{\circ }+2\angle QPR={180}^{\circ }$

$\therefore \angle QPR={45}^{\circ }$