ABCD is a parallelogram. If AB = 2AD and P is the mid-point of CD; prove that : ∠APB = 90°.

Given: ABCD is a parallelogram. AB = 2AD and P is the mid point of CD.

To prove: ∠APB = 90°

Proof: Since ABCD is a parallelogram, the opposite sides of the parallelogram are equal.

⇒ AB = DC and AD = BC

AB = 2AD ⇒ AD = $\dfrac{1}{2}$ AB and BC = $\dfrac{1}{2}$ AB ……………(1)

Since P is the mid-point of CD and AB = DC, we have:

DP = PC = $\dfrac{1}{2}$ DC = $\dfrac{1}{2}$ AB ……………(2)

From equations (1) and (2), we conclude that AD = DP. In Δ APD, since two sides are equal, the angles opposite these sides are equal:

∠APD = ∠DAP ………………(3)

Since, AB ∥ DC and AP is a transversal, the alternate interior angles are equal:

∠APD = ∠PAB ……………(4)

From equation (3) and (4), we get:

⇒ ∠DAP = ∠PAB

Let ∠DAP = ∠PAB = $\dfrac{1}{2}$ ∠A ……………(5)

Similarly, from equations (1) and (2), BC = PC. In Δ BPC, since two sides are equal, the angles opposite to equal sides are equal:

∠CPB = ∠PBC …………….(6)

Since AB ∥ Dc and BP is a transversal, the alternate interior angles are equal:

∠CPB = ∠PBA ……………(7)

From equations (6) and (7), we get:

⇒ ∠PBC = ∠PBA

Let ∠PBC = ∠PBA = $\dfrac{1}{2}$ ∠B ……………(8)

In a parallelogram, adjacent angles are supplementary:

∠A + ∠B = 180°

⇒ $\dfrac{1}{2}$ ∠A + $\dfrac{1}{2}$ ∠B = $\dfrac{1}{2}$ 180°

⇒ $\dfrac{1}{2}$ ∠A + $\dfrac{1}{2}$ ∠B = 90°

In Δ APB, sum of all angles is 180°.

⇒ ∠APB + ∠ABP + ∠BAP = 180°

⇒ ∠APB + $\dfrac{1}{2}$ ∠B + $\dfrac{1}{2}$ ∠A = 180°

⇒ ∠APB + 90° = 180°

⇒ ∠APB = 180° - 90°

⇒ ∠APB = 90°

Hence, ∠APB = 90°