Question 1166450
<pre>
We mark the line off with 32 marks, with A at 0 and B at 32.

{{{drawing(700,80,-.2,32.3,-40,40, graph(700,80,-.2,32.3,-40,40),
locate(-.24,20,A),locate(32-.2,20,B) 

 )}}}


C is the midpoint of AB​,

So we mark C at the 16 marker.

{{{drawing(700,80,-.2,32.3,-40,40, graph(700,80,-.2,32.3,-40,40),
locate(-.24,20,A),locate(32-.2,20,B),locate(16-.2,20,C) )}}} 

D is the midpoint of AC,

So we mark D at the 8 marker.

{{{drawing(700,80,-.2,32.3,-40,40, graph(700,80,-.2,32.3,-40,40),
locate(-.24,20,A),locate(32-.2,20,B),locate(16-.2,20,C),
locate(8-.2,20,D)
 )}}} 


E is the midpoint of AD,

So we mark E at the 4 marker.

{{{drawing(700,80,-.2,32.3,-40,40, graph(700,80,-.2,32.3,-40,40),
locate(-.24,20,A),locate(32-.2,20,B),locate(16-.2,20,C),
locate(8-.2,20,D), locate(4-.2,20,E)
 )}}} 

F is the midpoint of ED,

So we mark F at the 6 marker.

{{{drawing(700,80,-.2,32.3,-40,40, graph(700,80,-.2,32.3,-40,40),
locate(-.24,20,A),locate(32-.2,20,B),locate(16-.2,20,C),
locate(8-.2,20,D), locate(4-.2,20,E), locate(6-.2,20,F)
 )}}} 

G is the midpoint of EF,

So we mark G at the 5 marker.

{{{drawing(700,80,-.2,32.3,-40,40, graph(700,80,-.2,32.3,-40,40),
locate(-.24,20,A),locate(32-.2,20,B),locate(16-.2,20,C),
locate(8-.2,20,D),  locate(4-.2,20,E), locate(6-.2,20,F),locate(5-.2,20,G)
 )}}} 

H is the midpoint of DB,

We observe that D is at 8, and that B is at 32. So their midpoint is at
their average, (8+32)/2 = 20  

So we mark H at the 20 marker.

{{{drawing(700,80,-.2,32.3,-40,40, graph(700,80,-.2,32.3,-40,40),
locate(-.24,20,A),locate(32-.2,20,B),locate(16-.2,20,C),
locate(8-.2,20,D)  locate(4-.2,20,E), locate(6-.2,20,F),locate(5-.2,20,G),
locate(20-.2,20,H)
 )}}} 

If DC=56​, what is​ GH?

D is at 8 and C is at 16.  That's 16-8=8 units on our scale.  Since DC=56, 
Each of the units on our scale here counts 56/8=7 on the scale of the
problem. 

G is at 5, and H is at 20.  That's 20-5=15 units on our scale, and since
each unit on our scale is 7 units on the scale of the problem, the answer is
15∙7 = 105 on the scale of the problem.   

Edwin</pre>