Stumped by (simple) geometry problem

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I've been quite under desperate lately. I got no clue on how to solve this (apparently easy) geometry problem:

Consider a circle C of center O in a plane and a line segment AB not necessarily outside the circle C.
Having no compass, just a pen and a simple ruler, construct 2 points M and N inside the segment AB so that

$$ \frac{AM}{2,333...} = \frac{MN}{1,666...} = \frac{NB}{2} $$

Any ideas?
 
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Note that:- ##MN=\frac{5}{6}NB\\
AM=\frac{7}{6}NB\\
AB=3NB##
Now draw a long line ##BY## from ##B##
Now match one of the endpoints of the scale with the center of the given circle, and using the pen mark the radius of the circle onto the scale.
Now use this distance and mark 18 equidistant points on the line ##BY##. Now join the 18th point with ##A##. Slide the scale along ##BY## and join the remaining points with ##AB## as well. (sliding will make sure all the lines are parallel.)
So now we have divided ##AB## into 18 equal parts. The 6th point will be ##N## and the 11th point along the same direction will be ##M##.
fig-4-6.png
 
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certainly said:
Note that:- ##MN=\frac{5}{6}NB\\
AM=\frac{7}{6}NB\\
AB=3NB##
Now draw a long line ##BY## from ##B##
Now match one of the endpoints of the scale with the center of the given circle, and using the pen mark the radius of the circle onto the scale.
Now use this distance and mark 18 equidistant points on the line ##BY##. Now join the 18th point with ##A##. Slide the scale along ##BY## and join the remaining points with ##AB## as well. (sliding will make sure all the lines are parallel.)
So now we have divided ##AB## into 18 equal parts. The 6th point will be ##N## and the 11th point along the same direction will be ##M##.
View attachment 83272
Is there a way to figure that out from some theorems? Or its something you either should read in some old paper or try and fail until you find out?
 
Shyan said:
Is there a way to figure that out from some theorems? Or its something you either should read in some old paper
Not really. It's mostly logic and practice :-)
[But you could know before-hand the way to divide a line into equal parts. But it's simple enough that it's not necessary.]