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 given the following conditions in the problem:

- $\overline{AB} \parallel \overline{CD}$,
- $EF = FG$,
- $\angle AEG = x^\circ$,
- $\angle BEF = 100^\circ + x^\circ$.

Our goal is to find the value of $x$.

### Step 1: Analyze the angles
From the diagram, we know that lines $\overline{AB}$ and $\overline{CD}$ are parallel, and point $E$ lies on line $\overline{AB}$. Point $F$ lies on line $\overline{CD}$, and points $F$ and $G$ lie along the transversal $EG$. 

Thus, since $\overline{AB} \parallel \overline{CD}$, angles that are formed by the transversal are related. Specifically, we observe the angles at $E$ and $F$.

### Step 2: Use the given angle information
We are told that:
- $\angle AEG = x^\circ$,
- $\angle BEF = 100^\circ + x^\circ$.

### Step 3: Consider the external angles
Since $EF = FG$, triangle $EFG$ is isosceles. This means that the base angles, $\angle BEF$ and $\angle GEF$, are equal. Thus, we have:
\[
\angle GEF = 100^\circ + x^\circ.
\]

### Step 4: Sum of angles in a triangle
Now, consider triangle $AEG$. The sum of the angles in any triangle must be $180^\circ$. Therefore, we can write the equation for triangle $AEG$ as:
\[
\angle AEG + \angle GEF + \angle EAG = 180^\circ.
\]
Substitute the known values:
\[
x^\circ + (100^\circ + x^\circ) + \angle EAG = 180^\circ.
\]
Simplifying:
\[
2x^\circ + 100^\circ + \angle EAG = 180^\circ.
\]
This simplifies to:
\[
2x + \angle EAG = 80^\circ.
\]

### Step 5: Solve for $x$
Since $\angle EAG$ must be a positive angle, we now solve this equation to find the value of $x$.

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 $x$.

This problem involves finding the value of $x$ given that $\overline{AB} \parallel \overline{CD}$, $EF = FG$, and angles in a triangle with parallel lines and transversal. Learn how to apply the Isosceles Triangle Theorem and the sum of angles in a triangle to find the solution. Suitable for students in Grades 9-10.

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