Before I state my problem description, would like to describe problem which was stated before my problem. So it is like this

Given a line segment $AB$. You are required to divide it internally in the ratio $2 : 3$. steps for this problem is following

  1. Draw a ray $AC$ making an acute angle with $AB$.

  2. Starting with $A$, mark off 5 points $C_1, C_2, C_3, C_4, C_5$ at equal distances from the point $A$.

  3. Join $C_5$ and $B$
  4. Through $C_2$ (i.e. the second point), draw $C_2D$ parallel to $C_5B$ meeting $AB$ in $D$.

Then $D$ is required point, which divides $AB$ segment into $2:3$ part.

Here is picture which demonstrate this procedure: enter image description here

based on this information, I am trying to solve similar one

Draw a line segment 7 cm long. Divide it internally in the ratio $3 : 4$. Measure each part. Also write the steps of construction.

So as I understood, because we had to divide into ratio $2:3$ and we took $5$ points,in our case we have to take $7$ points right? Because of $3:4$ ration, with equal distance from starting $A$, of course first we should draw ray with acute angle,then took $7$ point,connect last point to one of the end of our segment (which equal to $7$cm) then choose $C_3$ and connect segment, intersection point would be desired point yes? Please tell me if I am wrong, also what should be length of each small segments, by which we divide large one?

  • $\begingroup$ yes it is right $\endgroup$ Apr 25, 2012 at 11:29
  • $\begingroup$ By small segments you mean those with the seven point (and A) as endpoints? It does not matter what their length is, anything constructable. Just make them the same. $\endgroup$ Apr 25, 2012 at 12:16
  • $\begingroup$ so my argument is correct? $\endgroup$ Apr 25, 2012 at 13:25
  • 1
    $\begingroup$ @dato: Yes, your method is correct for the ratio $3:4$, and generalizes in a straightforward way to the ratio $m:n$, where $m$ and $n$ are positive integers. It all comes down to properties of parallel lines (you get equal angles, so similar triangles), or to put it another way, properties of scaling. $\endgroup$ Apr 25, 2012 at 14:36
  • $\begingroup$ thanks very much @André Nicolas $\endgroup$ Apr 26, 2012 at 6:27


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