Given two perpendicular bisectors and a vertex, trace the triangle.

Question

Given two lines $l_1$ and $l_2$ that are not parallel and a point $P$ that is exterior to both of them, trace the triangle with one vertex on $P$ that has $l_1$ and $l_2$ as perpendicular bisectors of two of its sides.

My solution is as follows:

Take $P$ and mirror it from $l_1$, obtaining $P'$. Then mirror $P$ with respect to $l_2$, obtaining $P''$. Then, the triangle $PP'P''$ is a solution.

If I'm correct, there should be other 2 solutions: if $P'_1$ is now the reflection of $P'$ with respect to $l_2$, then another triangle is obtained.

The third solution is obtained when we mirror $P''$ with respect to $l_1$, in order to form the triangle $P'''P''P$. As we can see in the image, three non-congruent triangles can share the same circumcenter and the same vertex $P$

But my professor and classmates say there should only be one solution! Where exactly is my mistake?

Answer 1

Since we are dealing with perpendicular lines, if $l_1,l_2$ are perpendicular then there is a unique rectangular configuration with four coinciding possibilities.

If they are not perpendicular then yes there are four different possibilities, including the three you have already indicated.

Did you show your sketch here to your professor and classmates?

Answer 2

I really like your solution.

You not only found a solution with the vertex of p being opposite to the side for which the perpendicular bisector is not given, but two more solutions as well.

They just wanted the first triangle but the wording should have been clear regarding this condition.