Question

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

• A. $90°$
• B. $60°$
• C. $45°$
• D. $135°$

Answer 1

The correct option is A $90°$

$AD=BC$ [Opposite sides of a parllelogram]
$\Rightarrow AD=BC=\frac{1}{2}AB$........(1)
$AP=PB$ [P is midpoint of AB]........(2)
So, $AD=BC=AP=PB$
Thus, triangle $DAP$ is an isosceles triangle.
Therefore, $\angle ADP=\angle DPA$
Similarly, triangle $CPB$ is an isosceles triangle.
And $\angle CPB=\angle BCP$.
Now, $\angle PDC=\angle DPA$ [Alternate interior angles]
And $\angle DCP=\angle CPB$ [Alternate interior angles]
So, DP is the bisector of angle D and CP is the bisector of angle C.
We know that when bisectors of co-interior angles are drawn on the same side of the transversal, they are perpendicular to each other.
So, DP is perpendicular to CP.
Therefore, $\angle DPC=90°$.
Thus, the correct answer is option (a).