## Division in given ratio

Given three lines {a,b,c} intersecting at a point O and another point P, draw from P a line (e) which is divided in two  parts (x,y) by the three given lines, so that their ratio x/y = k for a given constant k.

Select on an arbitrary segment AC a point D dividing it into the given ratio k. Then draw arcs viewing these segment under the angles (a,b) and (b,c) respectively. The arcs intersect at a point B and define a triangle ABC and its secant BD so that the angles are (BA,BD)=(a,b) and (BD,BC)=(b,c). Draw from an arbitrary point X of line a triangle OXY similar to ABC.
Then draw from P a parallel to XY. This is the solution.
A simpler solution exists for the case k=-1 i.e. when line b defines the middle of the segment. It suffices then to take one arbitrary point Q on b and draw from there the parallels to a and b defining a parallelogram with one diagonal OQ. The other diagonal is the desired direction to which a parallel from P defines the secant solving the problem.