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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!

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

1 Answer 1

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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.

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