Question

If $P$ is mid-point of side $BC$ of a parallegram $ABCD$ such that $\angle BAP=\angle DAP$ if $AD=10$ cm then length of $CD$ is:

• A. $10$ cm
• B. $5$ cm
• C. $6$ cm
• D. $8$ cm

Answer 1

The correct option is B $5$ cm

Given $:$ ABCD is a parallelogram.

P is the mid point of BC.

$\angle BAP=\angle DAP$

ABCD is parallelogram

$\therefore AD||BC$

Also,

$\angle A+\angle B={180}^{\circ }$ ($\because$ Sum of adjacent interior angles in ${180}^{\circ }$) .... $\left(1\right)$

As given $\angle BAP=\angle DAP$

$\Rightarrow \angle BAP=\frac{1}{2}\angle A$ .... $\left(2\right)$

Now, in $△ABP:-$

$\angle 1+\angle 2+\angle B={180}^{\circ }$

Using $\left(1\right)$ and $\left(2\right):-$

$\Rightarrow \frac{1}{2}\angle A+\angle 2+{180}^{\circ }-\angle A={180}^{\circ }$

$\Rightarrow \angle 2=\frac{1}{2}\angle A$ $\to ⟨3⟩$

From $\left(2\right)$ and $\left(3\right)$ we see$:-$

$\angle 1=\angle 2$

$\therefore$ In $△ABP:-$

$\angle 1=\angle 2$

$\therefore BP=AB$ ($\because$ In a triangle, equal angles have equal sides opposite to them)

$\Rightarrow BC=2AB$

$\Rightarrow AD=2CD$ ($\because$ Opposites sides of parallelogram are equal.)

$\Rightarrow CD=5cm$