Question

Question 7
In the figure, DE || QR and AP and BP are bisectors of $\angle EABand\angle RBA$, respectively. Find $\angle APB$.

Answer 1

Given, DE || QR and AP and PB are the bisectors of $\angle EABand\angle RBA$ respectively.
The interior angles on the same side of transversal are supplementary.
$\therefore \angle EAB+\angle RBA={180}^{\circ }$
$\Rightarrow \frac{1}{2}\angle EAB+\frac{1}{2}\angle RBA=\frac{{180}^{\circ }}{2}$ [dividing both sides by 2]
$\Rightarrow \frac{1}{2}\angle EAB+\frac{1}{2}\angle RBA={90}^{\circ }$ . . . . . .(i)
Since AP and BP are the bisectors of $\angle EABand\angle RBA$ respectively,
$\therefore \angle BAP=\frac{1}{2}\angle EAB$ . . .. .(ii)
and $\angle ABP=\frac{1}{2}\angle RBA$ . . . . . (iii)
On adding equations (ii) and (iii), we get,
$\angle BAP+\angle ABP=\frac{1}{2}\angle EAB+\frac{1}{2}\angle RBA$
From eq. (i),
$\Rightarrow \angle BAP+\angle ABP={90}^{\circ }$ . . . (iv)
$In\Delta APB,\angle BAP+\angle ABP+\angle APB={180}^{\circ }$ [sum of all angles of a triangle is ${180}^{\circ }$]
$\Rightarrow {90}^{\circ }+\angle APB={180}^{\circ }$ [from eq. (iv)]
$\Rightarrow \angle APB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$