Question

State true or false:

In a square $ABCD$, diagonals meet at $O$. $P$ is a point on $BC$ such that $OB=BP$, then

$\angle POC={\left(22\frac{1}{2}\right)}^0$

• A. True
• B. False

Answer 1

The correct option is A True

Given a Square ABCD whose diagonals intersect at O.

All angles of a square are right angles & a diagonal of a square bisects the angles.

So $\angle ABC={90}^o\&\angle OBP=\frac{{90}^o}{2}={45}^o$

Also given OB=BP.

In isosceles $△$ BOP, $\angle BOP=\angle BPO$[ Base angles of a isosceles triangle are equal ]

So in $△$ BOP, $\angle OBP+\angle BOP+\angle BPO={180}^o$

$\Rightarrow 2\angle POB={180}^o-{45}^o={135}^o$

$\Rightarrow \angle POB=\frac{135}{2}$

$\Rightarrow \angle POB={67.5}^o$

We know that in a square diagonals are perpendicular bisectors of each other.

So $\angle COB={90}^o\Rightarrow \angle POC+\angle POB={90}^o\Rightarrow \angle POC+{67.5}^o={90}^o\Rightarrow \angle POC={90}^o-{67.5}^o\Rightarrow \angle POC={22.5}^o$