# Construct Triangle given bisectors and circumcircle

Suppose we have three concurrent lines $$g,h,k$$ in the Euclidean plane which meet at a point $$P\in g\cap h\cap k.$$ Moreover, let $$K$$ be some circle with center $$P$$ and some radius $$r>0$$. I would like to construct, with ruler and compass, all triangles with circumcircle $$K$$ and such that $$g,h,k$$ become the perpendicular bisectors of the sides of the triangle.

My ideas so far: I proved that the composition of reflections $$s_g\circ s_h\circ s_k$$ is again a reflection in some line through $$P$$ (where $$s_g$$ denotes the reflection in the line $$g$$ etc.). If we permute the order of the three reflections ($$s_g,s_h, s_k$$), we get again a reflection but in a different line through $$P$$.

Using this observation, I came up with the following idea for a construction. Suppose, we reflect some point $$A\in K\cap g$$ successively at $$k$$ and then at $$h$$. We obtain some point $$A'$$ (This would correspond to the point $$A'=s_h\circ s_k\circ s_g(A)$$, i.e. A' is the reflection of $$A$$ at some line $$d$$ through $$P$$). When we construct the bisector of $$A A'$$, we get the line of reflection $$d$$ of $$s_h\circ s_k\circ s_g$$. Let $$Q\in d\cap K$$. If we reflect the point $$Q$$ successively at $$g,k,h$$, we obtain a triangle, which meets all criteria.

If we do the same with all other points, we get in total two triangles which meet all criteria. But I do not know how to prove that there are no further such triangles. Moreover, it seems that my construction is somehow lengthy. Is it possible to construct the triangles more elegant?

I am very grateful for your help! Best wishes!

Your solution is great. If $$g,h,k$$ are bisectors of $$BC,CA,AB$$ respectively, since $$S_h\circ S_g\circ S_k(A)=A$$ and $$S_h\circ S_g\circ S_k(P)=P$$, and we know that $$S_h\circ S_g\circ S_k$$ is a line reflection $$S_d$$, we conclude that it must be $$d=PA$$. Thus $$A$$ must be an intersection of $$d$$ and the circumcircle, so there are only two possibilities for $$A$$, and hence only two solutions.

For another solution you can use the following facts: Let $$ABC$$ be a triangle, then:

Fact 1. Bisector of $$\angle A$$ and bisector of $$BC$$ intersect on the circumcircle.

Denote by $$A'$$ the point from Fact 1, and similarly define $$B'$$ and $$C'$$.

Fact 2. $$AA'\perp B'C'$$.

If $$A''$$ is another intersection of bisector of $$BC$$ with the circumcircle, and $$B''$$ and $$C''$$ are similarly defined.

Fact 3. On the circumcircle, the points are arranged as follows: $$A',B'',C',A'',B',C''$$.

Construction. Denote the intersections of $$g$$ with the cicumcircle by $$A',A''$$, of $$h$$ by $$B',B''$$ and of $$k$$ by $$C',C''$$, arranged as in Fact 3. By Fact 2, the heights in triangle $$A'B'C'$$ intersect the circumcircle in desired $$A,B,C$$. The second solution you get if you consider $$A''B''C''$$ instead of $$A'B'C'$$.