In the figure,$P$ is the mid-point of side $BC$ of a parallelogram $ABCD$ such that $\angle BAP = \angle DAP.$ Prove that $AD = 2 CD.$

(N/A) Given: $\angle BAP = \angle DAP = \frac{1}{2} \angle A \quad \dots(1)$
Since $ABCD$ is a parallelogram,we have $\angle A + \angle B = 180^{\circ} \quad \dots(2)$
[Sum of interior angles on the same side of the transversal is $180^{\circ}$]
In $\triangle ABP,$ we have $\angle BAP + \angle B + \angle APB = 180^{\circ}$
Substituting values from $(1)$ and $(2)$:
$\frac{1}{2} \angle A + (180^{\circ} - \angle A) + \angle APB = 180^{\circ}$
$\Rightarrow \angle APB = \frac{1}{2} \angle A \quad \dots(3)$
From $(1)$ and $(3),$ we get $\angle BAP = \angle APB.$
Therefore,$BP = AB$ [Sides opposite to equal angles are equal].
Since opposite sides of a parallelogram are equal,$AD = BC.$
$\Rightarrow \frac{1}{2} AD = \frac{1}{2} BC = BP$ [Since $P$ is the mid-point of $BC$].
$\Rightarrow \frac{1}{2} AD = AB$ [Since $BP = AB$].
Since $AB = CD$ (opposite sides of a parallelogram),we have:
$\frac{1}{2} AD = CD \Rightarrow AD = 2 CD.$
Hence,proved.