Question

In the adjoining figure, $ABCD$ is a square. if $\angle PQR={90}^o$ and $PB=QC=DR$. Prove that
$\angle QPR={45}^o$

Answer 1

We know that$PQ=QR$

Consider $△PQR$

From the figure we know that $\angle QPR$ and $\angle QRP$ are base angles of isosceles triangle

$\angle QPR=\angle QRP$

We know that the sum of all the angles in a triangle is ${180}^o$

$\angle QPR+\angle QRP+\angle PRQ={180}^o$

by substituting the values in the above equation

$\angle QPR+\angle QRP={180}^o-{90}^o$

$\angle QPR+\angle QRP={90}^o$

We know that $\angle QPR=\angle QRP$

so we get

$\angle QPR+\angle QPR={90}^o$

$2\angle QPR={90}^o$

$\angle QPR={45}^o$

Therefore it is proved that $\angle QPR={45}^o$