C is the midpoint of 

AB overbarAB​, 

D is the midpoint of 

AC overbarAC​, 

E is the midpoint of 

AD overbarAD​, 

F is the midpoint of 

ED overbarED​, 

G is the midpoint of 

EF overbarEF​, 

and H is the midpoint of 

DB overbarDB. 

If 

DCequals=88​, 

what is​ GH?

Let's approach this problem step by step.

We have a sequence of midpoints along a line segment \( AB \) with given properties:

1. \( C \) is the midpoint of \( AB \).
2. \( D \) is the midpoint of \( AC \).
3. \( E \) is the midpoint of \( AD \).
4. \( F \) is the midpoint of \( ED \).
5. \( G \) is the midpoint of \( EF \).
6. \( H \) is the midpoint of \( DB \).
7. \( DC = 8 \).

We are asked to find the length of \( GH \).

### Step 1: Determine the relationship between the segments

Let the length of \( AB \) be \( 2x \). Since \( C \) is the midpoint of \( AB \), we have:

\[ AC = CB = x \]

Now, since \( D \) is the midpoint of \( AC \):

\[ AD = DC = \frac{x}{2} \]

Given \( DC = 8 \), we know that:

\[ \frac{x}{2} = 8 \quad \Rightarrow \quad x = 16 \]

So, \( AB = 2x = 32 \).

### Step 2: Find the lengths of the other segments

- \( D \) is the midpoint of \( AC \), so \( AD = 8 \) (already given as \( DC = 8 \)).
- \( E \) is the midpoint of \( AD \), so:

\[ AE = ED = \frac{AD}{2} = \frac{8}{2} = 4 \]

- \( F \) is the midpoint of \( ED \), so:

\[ EF = FD = \frac{ED}{2} = \frac{4}{2} = 2 \]

- \( G \) is the midpoint of \( EF \), so:

\[ EG = GF = \frac{EF}{2} = \frac{2}{2} = 1 \]

- \( H \) is the midpoint of \( DB \), and since \( D \) and \( C \) are midpoints, \( H \) is \( \frac{3}{4} \) the way from \( D \) to \( B \):

\[ DB = 16 + 8 = 24 \]
\[ DH = HB = \frac{DB}{2} = \frac{24}{2} = 12 \]

### Step 3: Determine \( GH \)

Since \( GH \) is the difference between the positions of \( H \) and \( G \) on the line:

1. \( D \) is located 8 units from \( C \).
2. \( H \) is located 12 units from \( D \), so the total distance from \( C \) to \( H \) is \( 8 + 12 = 20 \) units.
3. \( G \) is located 9 units from \( C \) (since \( G \) is the midpoint between \( E \) and \( F \)).

Finally, the distance \( GH \) is:

\[ GH = 20 - 9 = 11 \]

So, \( GH = 11 \) units.

Would you like any further details or have additional questions?

### Here are five related questions to consider:

1. How does the concept of midpoints apply to the centroid of a triangle?
2. If the line segment \( AB \) were divided into more segments, how would you calculate the midpoint of any sub-segment?
3. What is the relationship between the distances in geometric progression when you keep taking midpoints?
4. How would the calculation change if \( C \) were not the midpoint of \( AB \)?
5. Can you derive the coordinates of \( H \) if the coordinates of \( A \) and \( B \) are given?

**Tip:** When working with midpoints, remember that each midpoint divides the segment into equal parts, so use the property that midpoints average the positions of their endpoints.

Explore how to calculate the length of GH in a sequence of midpoints on a line segment. Understand the concept of midpoints, segment division, and distance calculation. Suitable for advanced high school students.

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