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Let ABC be a triangle satisfying the condition for the question and let $\alpha$ be $\angle ABC$, Since $\triangle CHB$ is a right triangle we have $\angle HCB=90-\alpha$. Since we want the median and height to be trisectors all the angles at the top are $90^\circ-\alpha$.
Notice $\triangle CMB$ is an isosceles triangle since the height of that triangle is also the bisector. So if $HB$ measures $d$ then $MH$ measures $d$ also. So $AM$ must measure $2d$ since $M$ is the midpoint of $AB$.
$\triangle CAH$ is a right triangle and $CM$ is the bisector of $\angle ACM$. We want $\frac{MH}{AM}=\frac{1}{2}$. By the bisector theorem we have $\frac{MH}{AM}=\frac{CH}{CA}$. So we want $\frac{CH}{CA}=\cos(\angle ACH)=\frac{1}{2}$.
This means $\angle ACH=60^\circ\implies \angle ACB=90^\circ$
From here we also know $180-2\alpha=60^\circ\implies \alpha=60^\circ$
To sum things up we get $\angle ACB=90^\circ, \angle CBA=60^\circ, \angle BCA=30^\circ $