# In Fig. 8.7, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. Prove that AD = 2CD.

**Solution:**

Given, ABCD is a __parallelogram__

P is the midpoint of the side BC of the parallelogram

Given, ∠BAP = ∠DAP

We have to prove that AD = 2CD

We know that opposite sides of a parallelogram are parallel and __congruent__.

So, AD is parallel to BC

i.e., AD || BC ---------------------- (1)

Considering the two __parallel lines__ AD and BC cut by a transversal AP,

We know that the sum of __interior angles__ lying on the same side of the __transversal__ is always supplementary.

∠A + ∠B = 180°

∠B = 180° - ∠A -------------- (2)

Considering triangle ABP,

∠BAP + ∠B + ∠BPA = 180°

Since, ∠BAP = ∠DAP

∠BAP = ∠DAP = 1/2 ∠A

1/2 ∠A + ∠B + ∠BPA = 180°

From (2),

1/2 ∠A + 180° - ∠A + ∠BPA = 180°

∠BPA - 1/2 ∠A = 0

∠BPA = 1/2 ∠A

So, ∠BPA = ∠BAP

We know that the sides opposite to equal angles are equal.

So, AB = BP

Multiplying by 2 on both sides,

2AB = 2BP

Since P is the midpoint of BC

BP = CP

BC = BP + PC

BC = BP + BP

BC = 2BP

So, 2AB = BC

From (1),

2CD = AD

Therefore, it is proven that 2CD = AD

**✦ Try This: **In the adjoining figure, identify the pair of interior angles on the same side of the transversal.

**☛ Also Check:** NCERT Solutions for Class 9 Maths Chapter 8

**NCERT Exemplar Class 9 Maths Exercise 8.3 Problem 10**

## In Fig. 8.7, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. Prove that AD = 2CD.

**Summary:**

In Fig. 8.7, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. It is proven that AD = 2CD, since the opposite sides of a parallelogram are parallel and congruent

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