In this geometry exercise, students determine the value of angle x formed by intersecting lines, utilizing the properties of perpendicular lines and vertical angles.

Let's solve the problem step by step based on the image you provided.

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Problem:

We are given that:

- $ \overline{AB} \perp \overline{CD} $, meaning lines $ AB $ and $ CD $ are perpendicular.
- There is a point $ G $ where lines $ AB $, $ CD $, and another line $ EF $ intersect.
- An angle of $ 73^\circ $ is shown between line $ AB $ and line $ FG $ (specifically, between $ AG $ and $ FG $).
- We are to find the measure of angle $ x^\circ $, which is the angle between lines $ CG $ and $ EG $.

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Step 1: Understand the Geometry

We have:

- $ AB \perp CD $ → So the angle between $ AB $ and $ CD $ is $ 90^\circ $. This means:
- $ \angle AGC = 90^\circ $
- $ \angle BGC = 90^\circ $
- $ \angle AGD = 90^\circ $
- $ \angle BGD = 90^\circ $

Also, we see that line $ EF $ passes through point $ G $, forming angles with the other lines.

We're told that the angle between $ AG $ and $ FG $ is $ 73^\circ $. Since $ AB $ is horizontal and $ CD $ is vertical (because they are perpendicular), we can assume standard orientation.

Let’s label the angles carefully.

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Step 2: Use Vertical Angles and Supplementary Angles

Let’s focus on the angles around point $ G $.

The angle between $ AG $ and $ FG $ is $ 73^\circ $. That is, $ \angle AGF = 73^\circ $.

Now, since $ AB \perp CD $, and $ CD $ is vertical, then $ \angle AGC = 90^\circ $, as mentioned.

Now, consider triangle or angles formed at $ G $.

We want to find $ x^\circ $, which is the angle between $ CG $ and $ EG $, i.e., $ \angle CGE $.

But notice that $ EF $ is a straight line passing through $ G $, so $ \angle FGE = 180^\circ $.

So, let's look at angles along the vertical line $ CD $.

We know:

- $ \angle AGF = 73^\circ $
- $ \angle AGC = 90^\circ $

So, the angle between $ FG $ and $ CG $ is:

$$
\angle FGC = \angle AGC - \angle AGF = 90^\circ - 73^\circ = 17^\circ
$$

Now, $ \angle FGC $ and $ \angle EGC $ are vertical angles, because $ EF $ and $ CD $ intersect at $ G $. Wait — actually, $ \angle FGC $ and $ \angle EGD $ would be vertical angles. But we need to be careful.

Wait — let's think about the full picture.

Let’s define:

- $ \angle AGF = 73^\circ $ → this is below $ AB $, in the lower-left quadrant.
- Since $ AB \perp CD $, and $ CD $ is vertical, then $ CG $ is upward vertical, $ DG $ is downward vertical.

So, from point $ G $:

- $ AG $ is left horizontal
- $ BG $ is right horizontal
- $ CG $ is up vertical
- $ DG $ is down vertical
- $ FG $ goes down and to the left at $ 73^\circ $ from $ AG $
- $ EG $ goes up and to the right

Now, the angle between $ AG $ and $ FG $ is $ 73^\circ $. Since $ AG $ is horizontal and $ FG $ is going down-left, this suggests $ FG $ makes a $ 73^\circ $ angle with $ AG $, measured clockwise or counterclockwise?

Looking at the diagram, it seems the $ 73^\circ $ is the acute angle between $ AG $ and $ FG $, and since $ FG $ is below $ AB $, it's likely measured downward from $ AG $.

So, $ \angle AGF = 73^\circ $, and since $ \angle AGC = 90^\circ $ (from perpendicularity), then the angle between $ FG $ and $ CG $ is:

$$
\angle FGC = 90^\circ - 73^\circ = 17^\circ
$$

Now, $ \angle FGC $ and $ \angle EGD $ are vertical angles, but that might not help directly.

Instead, consider that $ \angle FGC = 17^\circ $, and $ \angle FGC $ and $ \angle EGC $ are adjacent angles along line $ EF $, but not necessarily supplementary unless we know more.

Wait — better idea:

Since $ EF $ is a straight line, the angles around $ G $ must sum appropriately.

But here's a key insight:

The angle between $ FG $ and $ CG $ is $ 17^\circ $, as we found.

Now, $ \angle CGE $ is the angle between $ CG $ and $ EG $. But $ EG $ is on the opposite side of $ FG $ on line $ EF $, so $ \angle CGE $ and $ \angle CGF $ are adjacent angles forming a straight line? No.

Wait — $ FG $ and $ EG $ are on the same line $ EF $, so $ \angle FGE = 180^\circ $.

So, if we go from $ FG $ to $ CG $, that’s $ 17^\circ $, and then from $ CG $ to $ EG $, that’s what we want — $ x^\circ $.

But $ CG $ is perpendicular to $ AB $, and $ EF $ is a line cutting through.

Actually, let's use vertical angles.

Notice that:

- The angle between $ FG $ and $ CG $ is $ 17^\circ $ (as calculated).
- Then, the angle between $ EG $ and $ CG $ should be equal to the angle between $ FG $ and $ DG $, due to symmetry?

Wait — better approach: use supplementary angles and vertical angles.

Let’s define:

- $ \angle AGF = 73^\circ $
- Since $ AB \perp CD $, $ \angle AGC = 90^\circ $
- Therefore, $ \angle FGC = 90^\circ - 73^\circ = 17^\circ $

Now, $ \angle FGC $ and $ \angle EGD $ are vertical angles, so they are equal.

But we want $ x^\circ = \angle CGE $

Note that $ \angle CGE $ and $ \angle FGC $ are not vertical.

Wait — let’s look at the angles around point $ G $.

We can consider triangle or just use linear pairs.

Alternatively, consider that:

- $ \angle AGF = 73^\circ $
- $ \angle AGC = 90^\circ $
- So $ \angle FGC = 17^\circ $

Now, $ \angle FGC $ and $ \angle EGD $ are vertical angles → both $ 17^\circ $

But we want $ \angle CGE $

Now, $ \angle CGE $ is the angle between $ CG $ and $ EG $

But $ \angle CGE $ and $ \angle DGF $ might be related?

Wait — let's try this:

Since $ EF $ is a straight line, and $ CD $ is a straight line, they intersect at $ G $, forming four angles.

We know:

- $ \angle FGC = 17^\circ $
- $ \angle FGC $ and $ \angle EGD $ are vertical angles → $ \angle EGD = 17^\circ $
- Now, $ \angle CGE $ and $ \angle FGD $ are vertical angles? Not exactly.

Wait — let’s list all angles at $ G $:

Lines $ AB $, $ CD $, and $ EF $ intersect at $ G $.

But $ AB \perp CD $, so they form four $ 90^\circ $ angles.

Line $ EF $ crosses them.

We know one angle: $ \angle AGF = 73^\circ $

This is the angle between $ AG $ and $ FG $

Then, since $ AB $ is horizontal and $ CD $ is vertical, the angle between $ FG $ and $ CG $ is $ 90^\circ - 73^\circ = 17^\circ $

So, $ \angle FGC = 17^\circ $

Now, $ \angle FGC $ and $ \angle EGD $ are vertical angles, so $ \angle EGD = 17^\circ $

Now, $ \angle CGE $ is adjacent to $ \angle EGD $ along the vertical line $ CD $? No.

Wait — $ \angle CGE $ is between $ CG $ and $ EG $

And $ \angle EGD $ is between $ EG $ and $ DG $

But $ CG $ and $ DG $ are opposite rays, so $ \angle CGD = 180^\circ $

So, $ \angle CGE + \angle EGD = 180^\circ $? No — only if $ E $ is on the line $ CD $, which it’s not.

Wait — better idea:

Consider triangle $ CGE $ — no, we don’t have a triangle.

Use the fact that $ \angle AGF = 73^\circ $, and $ AB \perp CD $, so $ \angle AGC = 90^\circ $

Then $ \angle FGC = 90^\circ - 73^\circ = 17^\circ $

Now, $ \angle FGC $ and $ \angle EGD $ are vertical angles → $ \angle EGD = 17^\circ $

Now, $ \angle CGE $ and $ \angle FGD $ are vertical angles? Let’s see:

- $ \angle CGE $: between $ CG $ and $ EG $
- $ \angle FGD $: between $ FG $ and $ DG $

Are these vertical? Only if $ CG $ and $ FG $ are opposite to $ EG $ and $ DG $, but they are not.

Wait — actually, when two lines intersect, vertical angles are opposite.

Here, lines $ CD $ and $ EF $ intersect at $ G $, so they form four angles:

- $ \angle CGE $
- $ \angle EGD $
- $ \angle DGF $
- $ \angle FGC $

Yes! These are the four angles formed by lines $ CD $ and $ EF $ intersecting at $ G $.

So:

- $ \angle FGC = 17^\circ $
- $ \angle EGD = 17^\circ $ (vertical to $ \angle FGC $)
- Then $ \angle CGE $ and $ \angle DGF $ are the other two angles, and they are vertical to each other.

Now, since the sum of angles around a point is $ 360^\circ $, and the two pairs of vertical angles are equal:

- $ \angle FGC = \angle EGD = 17^\circ $
- $ \angle CGE = \angle DGF = ? $

Now, the total around $ G $ is:

$$
\angle FGC + \angle CGE + \angle EGD + \angle DGF = 360^\circ
$$

But $ \angle FGC = \angle EGD = 17^\circ $, and $ \angle CGE = \angle DGF = x $

So:

$$
17^\circ + x + 17^\circ + x = 360^\circ \\
2x + 34^\circ = 360^\circ \\
2x = 326^\circ \\
x = 163^\circ
$$

Wait — that can't be right, because $ x $ is marked as a small angle in the diagram.

Look back — in the diagram, $ x^\circ $ is the angle between $ CG $ and $ EG $, and it looks like an acute angle.

But according to this, it’s $ 163^\circ $? That doesn’t make sense.

Wait — maybe I made a mistake.

Wait — no. Let's reconsider.

We have:

- $ \angle FGC = 17^\circ $
- $ \angle EGD = 17^\circ $
- $ \angle CGE $ and $ \angle DGF $ are the other two angles.

But $ \angle CGE $ and $ \angle DGF $ are vertical angles, so equal.

But also, adjacent angles sum to $ 180^\circ $.

For example, $ \angle FGC $ and $ \angle CGE $ are adjacent and form a straight line along $ EF $? No — $ EF $ is the line, so $ \angle FGC $ and $ \angle CGE $ are not on the same line.

Wait — actually, $ \angle FGC $ and $ \angle CGE $ are adjacent angles along line $ CD $? No.

Wait — better: consider the angles along line $ EF $.

On line $ EF $, the angles on one side must sum to $ 180^\circ $.

So, for example, $ \angle FGC $ and $ \angle CGE $ are adjacent angles along line $ EF $? No — $ CG $ is not on $ EF $.

Wait — the correct way:

Lines $ CD $ and $ EF $ intersect at $ G $, so they form four angles:

1. $ \angle CGE $ — between $ CG $ and $ EG $
2. $ \angle EGD $ — between $ EG $ and $ DG $
3. $ \angle DGF $ — between $ DG $ and $ FG $
4. $ \angle FGC $ — between $ FG $ and $ CG $

These four angles go around the point.

We know $ \angle FGC = 17^\circ $

Then $ \angle EGD = \angle FGC = 17^\circ $ (vertical angles)

Then $ \angle CGE $ and $ \angle DGF $ are the other pair of vertical angles.

Now, adjacent angles sum to $ 180^\circ $, so:

$ \angle FGC + \angle CGE = 180^\circ $? No — only if they are on a straight line.

But $ \angle FGC $ and $ \angle CGE $ are adjacent, but they share ray $ CG $, and their other rays are $ FG $ and $ EG $, which are on opposite sides of $ CG $.

But $ FG $ and $ EG $ are on the same line $ EF $, so yes — $ \angle FGC $ and $ \angle CGE $ are adjacent angles that together form the angle between $ FG $ and $ EG $, which is $ 180^\circ $, because $ FG $ and $ EG $ are opposite rays.

Ah! Yes!

Because $ EF $ is a straight line, the angle $ \angle FGE = 180^\circ $

Now, $ \angle FGE $ is composed of $ \angle FGC $ and $ \angle CGE $, but only if $ C $ is between $ F $ and $ E $ — which it’s not.

Wait — no. $ \angle FGE $ is the straight angle along $ EF $, so any path from $ F $ to $ E $ via $ G $ is $ 180^\circ $.

But $ \angle FGC $ is not part of $ \angle FGE $ — it involves $ C $, which is not on $ EF $.

So instead, consider the angles between $ EF $ and $ CD $.

We know $ \angle FGC = 17^\circ $

Then, since $ \angle FGC $ and $ \angle CGE $ are adjacent angles that together make the angle between $ FG $ and $ EG $, but only if $ CG $ is between them.

But $ CG $ is not between $ FG $ and $ EG $ — they are on different sides.

Wait — perhaps the best way is to realize that $ \angle AGF = 73^\circ $, and $ AB \perp CD $, so the angle between $ FG $ and $ CG $ is $ 90^\circ - 73^\circ = 17^\circ $

Now, this $ 17^\circ $ is the angle between $ FG $ and $ CG $

Then, since $ CG $ and $ EG $ are on opposite sides of $ EF $, the angle between $ CG $ and $ EG $ is the supplement of this angle?

No.

Wait — think about the angles around $ G $.

We have:

- $ \angle AGF = 73^\circ $
- $ \angle AGC = 90^\circ $
- So $ \angle FGC = 90^\circ - 73^\circ = 17^\circ $

Now, $ \angle FGC $ and $ \angle EGD $ are vertical angles → $ \angle EGD = 17^\circ $

Now, $ \angle CGE $ and $ \angle FGD $ are vertical angles.

Now, $ \angle FGD $ is the angle between $ FG $ and $ DG $

But $ \angle FGD $ and $ \angle AGF $ are adjacent along $ AB $? No.

Wait — consider triangle or use the fact that $ \angle AGF = 73^\circ $, and $ \angle AGC = 90^\circ $, so $ \angle FGC = 17^\circ $

Then, since $ \angle FGC = 17^\circ $, and $ \angle CGE $ is the angle between $ CG $ and $ EG $, and $ EG $ is on the opposite side of $ EF $, then the angle between $ CG $ and $ EG $ is the same as the angle between $ FG $ and $ DG $, because of vertical angles.

But we can calculate $ \angle CGE $ as follows:

From $ CG $, moving to $ EG $, we can go through the angle that is vertically opposite to $ \angle FGC $, but wait.

Actually, here’s the key:

The angle $ x^\circ $ is $ \angle CGE $

But $ \angle CGE $ and $ \angle FGD $ are vertical angles, so they are equal.

Now, $ \angle FGD $ is the angle between $ FG $ and $ DG $

Now, $ \angle AGF = 73^\circ $, and $ \angle AGD = 90^\circ $ (since $ AB \perp CD $), so $ \angle FGD = \angle AGD - \angle AGF = 90^\circ - 73^\circ = 17^\circ $

Wait — is that right?

$ \angle AGD = 90^\circ $, because $ AB \perp CD $, and $ AG $ is horizontal, $ DG $ is vertical down.

$ \angle AGF = 73^\circ $, which is the angle between $ AG $ and $ FG $

If $ FG $ is below $ AB $, and $ DG $ is down vertical, then the angle between $ FG $ and $ DG $ is $ \angle FGD = 90^\circ - 73^\circ = 17^\circ $

Yes!

So $ \angle FGD = 17^\circ $

But $ \angle FGD $ and $ \angle CGE $ are vertical angles, because $ CD $ and $ EF $ intersect at $ G $, and $ \angle FGD $ is between $ FG $ and $ DG $, while $ \angle CGE $ is between $ CG $ and $ EG $, and they are opposite.

So yes, $ \angle CGE = \angle FGD = 17^\circ $

Therefore, $ x = 17^\circ $

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✔ Final Answer:

$$
\boxed{17^\circ}
$$

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Explanation Summary:

- Given $ AB \perp CD $, so angles between them are $ 90^\circ $
- $ \angle AGF = 73^\circ $, so the angle between $ FG $ and $ CG $ is $ 90^\circ - 73^\circ = 17^\circ $
- But we want $ \angle CGE $, which is vertically opposite to $ \angle FGD $
- $ \angle FGD = 90^\circ - 73^\circ = 17^\circ $
- $ \angle CGE $ and $ \angle FGD $ are vertical angles → equal
- So $ x = 17^\circ $

✔ Answer: $ \boxed{17^\circ} $