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

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

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Answers (1)

Solution.
Given : ABCD is a parallelogram, P is a mid-point of BC such that $\angle BAP = \angle DAP$ .
To prove : $AD = 2CD$
Proof : Here ABCD is a parallelogram

$\therefore AD\parallel BC$ and AB is transversal
$\angle A+\angle B=180^{\circ}$ {sum of co-interior angles}
$\angle B=180-\angle A$    …..(1)
In $\triangle ABP$ ,
$\angle PAB+\angle BPA+\angle B=180^{\circ}$    {using angle sum property of triangle}
Putting the values,
$\frac{1}{2}\angle A+\angle BPA+(180^{\circ}-\angle A)=180^{\circ}$
$\angle BPA-\frac{\angle A}{2}=0$
$\angle BPA=\frac{\angle A}{2}$   …..(2)
$\Rightarrow \angle BPA=\angle BAP$
$AB = BP$    {opposite sides of equal triangle}
$2AB = 2BP$ {multiply both sides by 2}
$2AB = BC$    { P is midpoint of BC}
$2CD = AD$   { ABCD is parallelogram    $AB = CD$ and $BC = AD$ }
Hence proved