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

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

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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°