In the figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.
Given, $D E \| Q R$ and $A P$ and $P B$ are the bisectors of $\angle E A B$ and $\angle R B A$, respectively.
We know that, the interior angles on the same side of transversal are supplementary.
$\therefore$ $\angle E A B+\angle R B A=180^{\circ}$
$\Rightarrow \quad \frac{1}{2} \angle E A B+\frac{1}{2} \angle R B A=\frac{180^{\circ}}{2} \quad$ [dividing both sides by 2]
$\Rightarrow \quad \frac{1}{2} \angle E A B+\frac{1}{2} \angle R B A=90^{\circ}$ ...(i)
Since, $A P$ and $B P$ are the bisectors of $\angle E A B$ and $\angle R B A$, respectively.
$\therefore$ $\angle B A P=\frac{1}{2} \angle E A B$ .....(ii)
and $\angle A B P=\frac{1}{2} \angle R B A$ ...(iii)
On adding Eqs. (ii) and (iii), we get
$\angle B A P+\angle A B P=\frac{1}{2} \angle E A B+\frac{1}{2} \angle R B A$
From Eq. (i),
$\Rightarrow \quad \angle B A P+\angle A B P=90^{\circ}$ ..(iv)
In $\triangle A P B, \quad \angle B A P+\angle A B P+\angle A P B=180^{\circ}$ [sum of all angles of a triangle is $180^{\circ}$ ]
$\Rightarrow \quad 90^{\circ}+\angle A P B=180^{\circ}$ [from Eq. (iv)]
$\Rightarrow \quad \angle A P B=180^{\circ}-90^{\circ}=90^{\circ}$
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