# In the triangle $ABC$ $M$ is the middle of the side $AB$ and $CE$ is an altitude. Find the angles of triangle $ABC$ if...

In the triangle $$ABC$$ $$M$$ is the middle of the side $$AB$$ and $$CE$$ is an altitude. Find the angles of triangle $$ABC$$ if $$CM$$ and $$CE$$ split the angle $$ACB$$ in three equal parts.

I've figured out that triangles $$CME$$ and $$CBE$$ are congruent, but I'm not sure what to do exactly to be able to find out the exact angles of triangle $$ABC$$. ($$ACB=3x,ABC=90-x,CAB=270-2x$$).

Any help is appreciated, thanks!

• Please provide an accurate figure: If CE is an altitude, one should find a right angle in $E$ ! Jan 11, 2022 at 22:34
• A similar issue here Jan 11, 2022 at 23:34

If you know about $$\triangle CEM$$ and $$\triangle CEB$$, then you know about what kind of triangle is $$\triangle CEA$$. You might also know in what ratio $$E$$ splits $$MB$$ and as a consequence in what ratio $$M$$ splits $$AE$$. Now it's probably the time for angle bisector theorem, since $$CM$$ is a bisector of $$\triangle CEA$$. And in no time we received all the information we need to get the $$\angle A$$, and then other two angles.