Question

If two parallel lines are intersected by a transversal, then prove that bisectors of the interior angles from a rectangle.

Answer 1

AB and CD are two parallel line intersected by a transverse L

X and Y are the point of intersection of L with AB and CD respectively.

XP, XQ, YP and YQ are the angle bisector of $\angle AXY,\angle BXY,\angle CYXand\angle DYX$

$AB\parallel CD$ and L is transversal.

$\therefore \angle AXY=\angle DYX$ (pair of alternate angle)

$\Rightarrow \frac{1}{2}\angle AXY=\frac{1}{2}\angle DXY$

$\Rightarrow \angle 1=\angle 4\left(\angle 1=\frac{1}{2}\angle AXYand\angle 4=\frac{1}{2}\angle DXY\right)$

$\Rightarrow \frac{PX}{YQ}$

(If a transversal intersect two line in such a way that a pair of alternate

interior angle are equal, then the two line are parallel)

$\left(1\right)$

Also $\angle BXY=\angle CYX$ (pair of alternate angle)

$\Rightarrow \frac{1}{2}\angle BXY=\frac{1}{2}\angle CYX$

$\Rightarrow \angle 2=\angle 3\left(\angle 2=\frac{1}{2}\angle BXYand\angle 3=\frac{1}{2}\angle CYX\right)$

$\Rightarrow \frac{PY}{XQ}$

(If a transversal intersect two line in such a way that a pair of alternate

interior angle are angle, then two line are parallel)

$\left(2\right)$

from $\left(1\right)$ and $\left(2\right)$, we get

PXQY is parallelogram ....$\left(3\right)$

$\angle CYD={180}^0$

$\begin{array}{c}\Rightarrow \frac{1}{2}\angle CYD=\frac{180}{2}={90}^0\\ \Rightarrow \frac{1}{2}\left(\angle CYX+\angle DYX\right)={90}^0\\ \Rightarrow \frac{1}{2}\angle CYX+\frac{1}{2}\angle DYX={90}^0\\ \Rightarrow \angle 3+\angle 4={90}^0\\ \Rightarrow \angle PYQ={90}^0.......\left(4\right)\end{array}$

So using $\left(3\right)$ and $\left(4\right)$ we conclude that PXQY is a rectangle.

Hence proved.