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  • In Fig. 6.14, DE || QR and AP and BP are bisectors of ∠ EAB and ∠ RBA, respectively. Find ∠APB.

In Fig. 6.14, DE \parallel QR and AP and BP are bisectors of \angle EAB and \angle RBA, respectively. Find \angle APB.

Answers (1)

DE \parallel QR and AP and PB are the bisectors of \angle EAB and \angle RBA,
We know that the interior angles on the same side of the transversal are supplementary
So,
\angle EAB + \angle RBA = 180^{\circ}
\frac{1}{2} \angle EAB + \frac{1}{2}\angle RBA =\frac{1}{2} (180^{\circ})
\frac{1}{2}\angle EAB+\frac{1}{2}\angle RBA=90^{\circ} …(i)
AP and BP are the bisectors of \angle EAB and \angle RBA respectively.
\angle BAP=\frac{1}{2}\angle EAB                    …(ii)
\angle ABP = \frac{1}{2} \angle RBA                 …(iii)
On adding equations (ii) and (iii) we get
\angle BAP + \angle ABP = \frac{1}{2} \angle EAB +\frac{1}{2} \angle RBA
From equation (i)
\angle BAP + \angle ABP = 90^{\circ}
In \triangle APB,
\angle BAP + \angle ABP + \angle APB = 180^{\circ}
90^{\circ}+ \angle APB = 180^{\circ}
\angle APB = 180^{\circ} - 90^{\circ}= 90^{\circ}

Hence \angle APB = 90^{\circ}

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