Question

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

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

Answer 1

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

Given AB=2 AD and P is the mid point of AB then

AP=AD And PB=BC because opposite side of parallelogram are equal

In $\Delta DAP,\angle ADP=\angle DPA$

So $\angle ADP+\angle DPA+DAP={180}^0$

$\Rightarrow 2\angle ADP+\angle DAP=180$

$Similarlyin$\Delta PBC ,2\angleCPB+CBP=180^{0}$

In parallelogram ABCD $\angle DAP+\angle CBP={180}^0$

Then 2\angle ADP+\angle DAP+,2\angleCPB+CBP=180+180

Then $\angle DPA+\angle CPB={90}^0$

Then $\angle DPA+\angle CPB+\angle DPC={180}^0$

$\Rightarrow \angle DPC=180-90={90}^0$