Question

A line l is parallel to line m and a transversal p intersects them at X, Y respectively. Bisectors of interior angles at X and Y intersect at P and Q. Is PXQY a rectangle? Give reason.

Answer 1

Given, $l\parallel m$
$\therefore \angle DXY=\angle XYA$ [alternate interior angles]
$\Rightarrow \frac{\angle DXY}{2}=\frac{\angle XYA}{2}$ [dividing both the sides by 2]

Now, $\angle 1=\angle 2$ [$\because$ XP and YQ are bisectors]
Alternate angles are equal, i.e. $\angle 1=\angle 2$
$\Rightarrow XP\parallel QY$ ....... (i)
Similarly, $XQ\parallel PY$ ....... (ii)
From Eqs. (i) and (ii), we get
PXQY is a parallelogram ........ (iii)
$\Rightarrow \angle DXY+\angle XYB={180}^{\circ }$ [$\because$ interior angles on the same side of transversal are supplementary]
$\frac{\angle DXY}{2}+\frac{XYB}{2}=\frac{{180}^{\circ }}{2}$ [dividing both the sides by 2]
$\angle 1+\angle 3={90}^{\circ }$ ....... (iv)
In $\Delta XYP$,
$\angle 1+\angle 3+\angle P={180}^{\circ }$
${90}^{\circ }+\angle P={180}^{\circ }$ [from Eq. (iv)]
$\angle P={90}^{\circ }$ ......... (v)
Similarly, $\angle Q={90}^{\circ }$ ......... (vi)
From Eqs. (iii), (v) and (vi), PXQY is a rectangle.