Question

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

Answer 1

DE || QR (given)

∴∠EAB + ∠RBA = 180°

(Sum of the interior angles on the same side of a transversal is 180°)

⇒ \(\frac { 1 }{ 2 }\)∠EAB + \(\frac { 1 }{ 2 }\)∠RBA = 90° …(i)

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

\(\frac { 1 }{ 2 }\)∠EAB = ∠PAB

\(\frac { 1 }{ 2 }\)∠RBA = ∠PAB

∵In ∆APB

∠PAB + ∠PBA + ∠APB = 180°

⇒ 90° + ∠APB = 180°

⇒ ∠APB = 90°