# Construct a point with given ratio of distances to sides of an angle

We have an angle $$\angle AOB$$ and two segments of lengths $$x$$ and $$y$$. Construct a point $$P$$ inside $$\angle AOB$$ such that $$\frac{\mathrm{dist}(P, OA)}{\mathrm{dist}(P,OB)}=\frac{x}y.$$

My idea is to generalise the construction of the angle-bisector: we draw a circle with center $$O$$ intersecting $$OA$$ at $$A'$$ and $$OB$$ at $$B'$$ and then two circles with equal radiuses and centers at $$A'$$ and $$B'$$ and any of their intersection will be on the angle bisector. This might useful since this is the case $$x=y$$. Probably at some point we have to draw a circle with radius $$x$$ and one with radius $$y$$.

Another useful idea is given a segment of length $$a$$, the construction of a segment of length $$b$$ such that $$a/b=x/y$$ is easily done: let $$MN=x$$, $$P\in MN$$ such that $$NP=y$$ and $$XY\parallel MN$$ such that $$XY=a$$ (these can be easily constructed). Then we have to consider $$Q\in XM\cap YN$$ and $$Z\in QP\cap XY$$ and we will have $$YZ=b$$.