Question

ABCD is a parallelogram if \[AB = 2AD\] and P is the midpoint of AB then $\angle CPD = $
A) $45^\circ $
B) $270^\circ $
C) $180^\circ $
D) $90^\circ $

Answer 1

Hint:Use the properties of the parallelogram that the opposite sides are equal. The sum of the adjacent angles of the parallelogram is $180^\circ $. Use the properties of the isosceles triangle. The sum of the angles in the linear pair is $180^\circ $.

Complete step-by-step answer:
 Draw the diagram of the given parallelogram ABCD

We are given that ABCD is a parallelogram,\[AB = 2AD\] and P is the midpoint of AB.
We have to find the measure of $\angle CPD$.
$
  AB = 2AD \\
   \Rightarrow AD = \dfrac{{AB}}{2}....(1) \\
 $
Since, P is the midpoint of AB, therefore,
$AB = PB = \dfrac{{AB}}{2}....(2)$
From equation $(1)$ and $(2)$
$AD = AP = PB$
We know that opposite sides of the parallelogram are equal.
Therefore, $AD = BC$
Hence, $AD = AP = PB = BC$
Now we have to two different triangles to analyse first one is $\Delta APD$ and second one is $\Delta BPC$
First, we see $\Delta APD$
In $\Delta APD$
$AD = AP$
Since, two sides are equal, therefore, $\Delta APD$ is an isosceles triangle.
We know that the angle opposite to the equal sides are equal.
Therefore, $\angle ADP = \angle APD$
In $\Delta BPC$
$BC = PB$
Since, two sides are equal, therefore, $\Delta BPC$ is an isosceles triangle.
We know that the angle opposite to the equal sides are equal.
Therefore, $\angle BCP = \angle BPC$
Since, $AD = AP = PB = BC$, therefore, $\angle ADP = \angle APD = \angle BPC = \angle BCP$
Again, In $\Delta APD$
We know that the sum of the interior angles of the triangle is $180^\circ $
Therefore,
 \[
  \angle A + \angle APD + \angle ADP = 180^\circ \\
   \Rightarrow \angle A + \angle APD + \angle APD = 180^\circ \\
   \Rightarrow 2\angle APD = 180^\circ - \angle A \\
   \Rightarrow \angle APD = 90^\circ - \dfrac{{\angle A}}{2} \\
 \]
Similarly, $\angle BPC = 90^\circ - \dfrac{{\angle B}}{2}$
We know that the sum of the angles in the linear pair is $180^\circ $.
Therefore, $\angle APD + \angle BPC + \angle CPD = 180^\circ $
Substitute all the values.
$
  90^\circ - \dfrac{{\angle A}}{2} + 90^\circ - \dfrac{{\angle B}}{2} + \angle CPD = 180^\circ \\
   \Rightarrow 180^\circ - \dfrac{1}{2}\left( {\angle A + \angle B} \right) + \angle CPD = 180^\circ \\
   \Rightarrow \angle CPD = \dfrac{1}{2}\left( {\angle A + \angle B} \right)....(3) \\
 $
The sum of the adjacent angles of the parallelogram is $180^\circ $.
Therefore, $\angle A + \angle B = 180^\circ $
Substitute the value in equation $(3)$ and evaluate the value of $\angle CPD$.
$
  \angle CPD = \dfrac{{180^\circ }}{2} \\
   \Rightarrow \angle CPD = 90^\circ \\
 $
Hence, option (D) is correct.

Note:In these types of questions first draw the figure and analyse the conditions properly. The adjacent angles are those angles which have a common side. In the parallelogram ABCD, A and D, A and B are adjacent angles.