Question

Question 10
In figure, P is the mid-point of side BC of a parallelogram ABCD such that $\angle BAP=\angle DAP$. Prove that AD = 2CD .

Thinking process

Firstly, use the property that sum of cointeritor angles is ${180}^{\circ }$.Secondly use the property that sum of all angles in a triangle Is ${180}^{\circ }$ and then prove the required result.

Answer 1

Given in parallelogram ABCD, P is a mid-point of BC such that $\angle BAP=\angle DAP$

To prove AD = 2CD

Proof Since, ABCD is a parallelogram.

So, AD ∥ BC and AB is transversal, then

$\angle A+\angle B={180}^{\circ }$ [ Sum of cointerior angles is ${180}^{\circ }$]

$\Rightarrow \angle B={180}^{\circ }-\angle A$

In $\Delta ABP$, $\angle PAB+\angle B+\angle BPA={180}^{\circ }$ [by angle sum property of a triangle]

$\Rightarrow \frac{1}{2}\angle A+{180}^{\circ }-\angle A+\angle BPA={180}^{\circ }$ [ from eq.(i)]

$\Rightarrow \angle BPA-\frac{\angle A}{2}=0$

$\Rightarrow \angle BPA=\angle BAP$

$\Rightarrow AB=BP$ [ Opposite sides of equal angles are equal]

On Multiplying both sides by 2 we get

2AB = 2BP

$\Rightarrow 2AB=BC$ [Since P is the mid-point of BC]

$\Rightarrow$ [Since, ABCD is a parallelogram, then AB = CD and BC = AD]