Question

$ABCD$ is a parallelogram as shown in figure. If $AB=2AD$ and $P$ is mid-point of $AB$, then $\angle CPD=$

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

If $AB=2AD$, then,

$AD=\frac{1}{2}AB$ (1)

Also P is the midpoint of AB, then,

$AP=PB=\frac{1}{2}AB$ (2)

From equations (1) and (2), $AD=AP=PB=BC$.

So, $\angle ADP=\angle APD=\angle BCP=\angle BPC$. This shows that $\Delta ADP$ is an isosceles triangle.

$\angle A+\angle ADP+\angle APD={180}^{\circ }$

$\angle A+2\angle APD={180}^{\circ }$

$\angle APD={90}^{\circ }-\frac{\angle A}{2}$

In the same way, $\angle BPC={90}^{\circ }-\frac{\angle B}{2}$.

Also, $\angle APD+\angle BPC+\angle CPD={180}^{\circ }$

${90}^{\circ }-\frac{\angle A}{2}+{90}^{\circ }-\frac{\angle B}{2}+\angle CPD={180}^{\circ }$

$\angle CPD=\frac{\angle A+\angle B}{2}$

The sum of adjacent angles of a parallelogram is ${180}^{\circ }$, therefore,

$\angle CPD=\frac{{180}^{\circ }}{2}$

$\angle CPD={90}^{\circ }$