Question

In the following figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.

Answer 1

Given, DE || QR and AP and PB are the bisectors of ∠EAB and ∠RBA, respectively.

We know that, the interior angles on the same side of transversal are supplementary.

∴ ∠EAB + ∠RBA = 180°

⇒ `1/2 ∠ EAB + 1/2 ∠RBA = (180^circ)/2`  ...[Dividing both sides by 2]

⇒ `1/2 ∠EAB + 1/2 ∠RBA = 90^circ`   ...(i)

Since, AP and BP are the bisectors of ∠EAB and ∠RBA, respectively.

∴ `∠BAP = 1/2 ∠EAB`   ...(ii)

And `∠ABP = 1/2 ∠RBA`  ...(iii)

On adding equations (ii) and (iii), we get

`∠BAP + ∠ABP = 1/2 ∠EAB + 1/2 ∠RBA`

From equation (i),

⇒ ∠BAP + ∠ABP = 90°

In ΔAPB, ∠BAP + ∠ABP + ∠APB = 180°  ...[Sum of all angles of a triangle is 180°]

⇒ 90° + ∠APB = 180°  ...[From equation (iv)]

⇒ ∠APB = 180° – 90° = 90°