Question

In the given figure, $AB||CD$ and a transversal $EF$ cuts them at $G$ and $H$ respectively.
If $GL$ and $HM$ are the bisectors of the alternate angles $\angle AGH$ and $\angle GHD$ respectively, prove that $GL||HM$.

Answer 1

Given: $AB\parallel CD$, $GL$ and $HM$ are angle bisectors of $\angle AGH$ and $\angle GHD$ respectively.

Now,

$\angle AGH=\angle GHD$ (alternate interior angles as $AB\parallel CD$)

$\Rightarrow \frac{1}{2}\angle AGH=\frac{1}{2}\angle GHD$

$\Rightarrow \angle LGH=\angle GHM$

But they also form a pair of equal interior angles.

Hence, $GL\parallel HM$ [$\because$ if the angles of any pair of alternate interior angles are equal, then the lines are parallel]