91Ó°ÊÓ

As per the given information, $\stackrel{¯}{BF}$bisects $∠ABE$and $\stackrel{¯}{DG}$bisects $∠CDB$which implies that,

$\begin{array}{c}∠ABF=∠FBE\\ ∠CDG=∠GDB\end{array}$

If two parallel lines are bisected by a transversal then the corresponding angles are congruent.

Here,$\stackrel{¯}{AB}âˆ\yen \stackrel{¯}{CD}$ and$\stackrel{¯}{DE}$ is transversal then $∠CDXâ‰\dots ∠ABD$.

Now, $∠CDX+∠CDG+∠GDB=180$and $∠ABD+∠ABF+∠FBE=180$then from these two equations we get,

$\begin{array}{c}∠CDX+∠CDG+∠GDB=∠ABD+∠ABF+∠FBE\\ ∠CDX+∠CDG+∠CDG=∠ABD+∠ABF+∠ABF\\ ∠CDX+2∠CDG=∠ABD+2∠ABF\\ 2∠CDG=2∠ABF\\ ∠CDG=∠ABF\end{array}$

If two parallel lines are bisected by a transversal then the corresponding angles are congruent.

Here, $\stackrel{¯}{AB}âˆ\yen \stackrel{¯}{CD}$and$\stackrel{¯}{GD}$ is transversal then $∠CDGâ‰\dots ∠AYG$.

As $∠CDGâ‰\dots ∠ABF$and $∠CDGâ‰\dots ∠AYG$implies $∠ABFâ‰\dots ∠AYG$, which are corresponding angles. Therefore, $\stackrel{¯}{BF}âˆ\yen \stackrel{¯}{DG}$.

Hence it is proved that $\stackrel{¯}{BF}âˆ\yen \stackrel{¯}{DG}$.