We are given that $\overline{EJ} \parallel \overline{FK}$, $\overline{JG} \parallel \overline{KH}$, and $\overline{EF} \cong \overline{GH}$. We want to prove that $\triangle EJG \cong \triangle FKH$.

1. Problem Description

We are given that $\overline{EJ} \parallel \overline{FK}$, $\overline{JG} \parallel \overline{KH}$, and $\overline{EF} \cong \overline{GH}$. We want to prove that $\triangle EJG \cong \triangle FKH$.

2. Solution Steps

First, since $\overline{EF} \cong \overline{GH}$, we know that $EF = GH$. Add $FG$ to both sides of the equation:

$EF + FG = GH + FG$

$EG = FH$

Since $\overline{EJ} \parallel \overline{FK}$, $\angle JEF$ and $\angle KFE$ are consecutive interior angles formed by transversal $\overline{EF}$.

Also, $\angle JEG$ and $\angle KFH$ are corresponding angles, since $\overline{EJ} \parallel \overline{FK}$. These are not the same angles.

Since $\overline{EJ} \parallel \overline{FK}$, we know that $\angle JEG \cong \angle KFH$.

Since $\overline{JG} \parallel \overline{KH}$, we know that $\angle EGJ \cong \angle FHK$.

In $\triangle EJG$ and $\triangle FKH$, we have $EG = FH$, $\angle JEG \cong \angle KFH$ and $\angle EGJ \cong \angle FHK$. Therefore, by Angle-Side-Angle (ASA) congruence postulate, $\triangle EJG \cong \triangle FKH$.

Thus, given that $\overline{EJ} \parallel \overline{FK}$, $\overline{JG} \parallel \overline{KH}$, and $\overline{EF} \cong \overline{GH}$, we can prove $\triangle EJG \cong \triangle FKH$. Since $\overline{EF} \cong \overline{GH}$, we know that $EF = GH$. Then we have $EF + FG = GH + FG$, which gives us $EG = FH$.

Also, since $\overline{EJ} \parallel \overline{FK}$, we know that $\angle JEG \cong \angle KFH$, and since $\overline{JG} \parallel \overline{KH}$, we know that $\angle EGJ \cong \angle FHK$.

Therefore, by the ASA congruence postulate, $\triangle EJG \cong \triangle FKH$.

3. Final Answer

$\triangle EJG \cong \triangle FKH$.