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
Draw a ray $AC$ making an acute angle with $AB$.
Starting with $A$, mark off 5 points $C_1, C_2, C_3, C_4, C_5$ at equal distances from the point $A$.
- Join $C_5$ and $B$
- 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:
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?