Question

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

State True or False: $\angle BDC=2\angle POC$

• A. True
• B. False

Answer 1

The correct option is A True

In square ABCD,

$OB=BP$ (Given)

$\therefore \angle DPB=\angle BOP$ [Angular opposite to equal sides]

$\angle DPB=\frac{{90}^{\circ }}{2}={45}^{\circ }$

$\angle BOP=\frac{{180}^{\circ }-{45}^{\circ }}{2}={67.5}^{\circ }$

$\angle BOC={90}^{\circ }$ [Diagonal bisect at right angle]

$\angle POC={90}^{\circ }-{67.5}^{\circ }={22.5}^{\circ }$

$△BDC$ is an isosceles $△$ , with $\angle BDC={90}^{\circ }$ & $\angle BDC=\angle DBC={45}^{\circ }$

Therefore, $\angle BDC=2\angle POC$

Hence statement is true.