We are tasked with finding the value of $x$ given the conditions in the problem. The key information includes:

• $\overline{AB} \parallel \overline{CD}$, meaning that the lines $\overline{AB}$ and $\overline{CD}$ are parallel.
• $EF = FG$, indicating that segment $EF$ is congruent to segment $FG$.
• $\angle AEG = x^\circ$.
• $\angle BEF = 100^\circ + x^\circ$.

Step 1: Identify angles involving the parallel lines

Since $\overline{AB} \parallel \overline{CD}$, we can use properties of parallel lines and transversals. The transversal $EF$ intersects both parallel lines $\overline{AB}$ and $\overline{CD}$, creating corresponding and alternate interior angles.

Step 2: Use properties of isosceles triangle $EFG$

We are told that $EF = FG$, which makes triangle $EFG$ isosceles. In any isosceles triangle, the angles opposite the equal sides are congruent. Therefore, we know:

$\angle EFG = \angle FEG$

Let these angles be denoted by $y$. Hence:

$\angle EFG = \angle FEG = y$

Step 3: Sum of angles in triangle $EFG$

The sum of angles in any triangle is 180°. Applying this to triangle $EFG$, we get:

$\angle AEG + \angle EFG + \angle FEG = 180^\circ$

Substituting the known values:

$x + y + y = 180^\circ$

This simplifies to:

$x + 2y = 180^\circ \quad \text{(1)}$

Step 4: Analyze $\angle BEF$

We are given that $\angle BEF = 100^\circ + x^\circ$. Since $EF = FG$, $\angle EFG = y$, and by the isosceles triangle property, we know that the angle at vertex $F$ (opposite the base $EG$) must fit into the overall angle structure. Considering