This is part of a bigger problem. We are given three points $A,B,C$ in general position. The goal is to construct a triangle $AEF$ such that the segment $AB$ is a median and $C$ is the orthogonal projection of $E$ on $AF$.
We trace the line $AC$, we know $F$ will lie on this line. We can then trace the altitude by constructing the perpendicular to $AC$ passing through $C$, we know $E$ will lie on this new line.
But I am now stuck with two lines and a point $B$. I need to find a segment whose endpoints lie on each line and whose middle is $B$. I have tried for quite a while and am not sure if this is possible.
Does anybody has an idea? Thanks in advance!
Hint: median to hypotenuse of a right triangle is half of hypotenuse. Consider right triangle $CEF$ and median $CB$. If you can find $CB$ then $EF=2CB$.