In figure 5,35 DE || QR and AP and BP are respectively the bisectors of ∠EAB and ∠RBA Find the value of ∠APB.

Given:

DE || QR

AP and BP are respectively the bisectors of ∠EAB and ∠RBA

Theory:

⇒ If 2 lines are parallel then sum of their internal angles is 180°

⇒ Sum of the angle of a triangle is 180°

As DE || QR

And n is the transverse

By interior angles

⇒ ∠ EAB + ∠ RBA = 180°

If AP and BP are respectively the bisectors of ∠EAB and ∠RBA

⇒ ∠ PAB =

⇒ ∠ PBA =

By adding both we get;

⇒ ∠ PAB + ∠ PBA = = = = 90°

In Δ APB

∠ PAB + ∠ PBA + ∠ APB = 180°

90° + ∠ APB = 180°

⇒ ∠ APB = 180° - 90° = 90°