Triangle given its three perpendicular bisectors and a point of an edge

Question

Is it possible to determine a triangle given its three perpendicular bisectors (meeting at a point which will be the circumcenter) and, say, a point of an edge, or any condition that can make the solution unique, using compass and straightedge? Of course I could put a system of equations, but I'm looking for a graphical procedure.

Answer 1

Yes, it seems like the question you have in mind is enough.

Suppose $P$ is the given point and $l,m,n$ are the perpendicular bisectors. Also assume that it is known that $P$ lies on the side which is perpendicular to $l$.

Now draw a line $q$ which is perpendicular to $l$, passing through $P$. One side of the triangle lies along this line.

Now reflect line $q$ on $m$ to give a new line $q_m$.

Similar reflect line $q$ on $n$ to give a new line $q_n$.

The intersection point of $q_m$ and $q_n$ is a vertex of the triangle and we can construct the whole triangle, given that point.