In quadrilateral ABCD, ∠D = 40°, ∠C = 80° AP bisects angle A and BP bisects angle ∠B; then ∠APB is equal:

1. 90°

2. 120°

3. 90°

4. 60°

As we know, the sum of all the angles in a quadrilateral is 360°.

In quadrilateral ABCD, ∠D = 40°, ∠C = 80°.

∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠A + ∠B + 80° + 40° = 360°

⇒ ∠A + ∠B + 120° = 360°

⇒ ∠A + ∠B = 360° - 120°

⇒ ∠A + ∠B = 240°

⇒ $\dfrac{1}{2}$(∠A + ∠B) = $\dfrac{1}{2}$ 240°

⇒ $\dfrac{1}{2}$(∠A + ∠B) = $\dfrac{240°}{2}$

⇒ $\dfrac{1}{2}$(∠A + ∠B) = 120°

As we know, the sum of all the angles in a triangle is 180°.

∠PAB + ∠PBA + ∠APB = 180°

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

⇒ 120° + ∠APB = 180°

⇒ ∠APB = 180° - 120°

⇒ ∠APB = 60°

Hence, option 4 is the correct option.