In the diagram below, we have $\overline{AB}\parallel\overline{CD}$, $EF = FG$, $\angle AEG = x^\circ$, and $\angle BEF = 100^\circ + x^\circ$. Find the value of $x$.

[asy]
pair A,B,C,D,EE,F,G;
D = (0,0);
C = (1.5,0);
A = (1.5,0.7);
B = (0,0.7);
draw (A--B);
draw(C--D);
EE = (0.2,0.7);
F = (0.4,0);
G = F + (sqrt(0.53),0);
draw(F--EE--G);
label("$G$",G,S);
label("$F$",F,S);
label("$E$",EE,N);
label("$A$",A,E);
label("$C$",C,E);
label("$B$",B,W);
label("$D$",D,W);
dot(A);
dot(B);
dot(C);
dot(D);
[/asy]

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

In the diagram below, we have $\overline{AB}\parallel\overline{CD}$, $EF = FG$, $\angle AEG = x^\circ$, and $\angle BEF = 100^\circ + x^\circ$. Find the value of $x$.

Solve the problem involving parallel lines, an isosceles triangle, and given angle relationships. Key geometry concepts include the sum of angles in a triangle and properties of parallel lines. Suitable for students in Grades 9-11.

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