Show that a triangle is right angled In $\triangle ABC$ (not isosceles) $CH$, $CL$ and $CM$ are respectively height, bisector and median. Show that $\angle ACB = 90^\circ$ if and only if $\angle HCL = \angle MCL$.
I think that I have to show that $\triangle MCB$ (or $\triangle ACM$) is isosceles, but I can't figure it out. I will be very grateful if you give me a hint.


If $\angle ACB=90^\circ$, then $\angle BCM=\angle ABC$, $\angle BCL=45^\circ$ and $\angle BCH = 90^\circ-\angle ABC$. So, we have $\angle HCL=\angle ABC-45^\circ=\angle MCL$.
Conversely, if $\angle HCL=\angle MCL$, then $\angle ACM=\angle BCH$ and $\angle BCM=\angle ACH$.
$\displaystyle \frac{AC}{AM}=\frac{\sin\angle AMC}{\sin\angle ACM}=\frac{\sin \angle BMC}{\sin \angle BCH}=\frac{\sin \angle BMC}{\sin \angle BCM}\cdot\frac{\sin \angle ACH}{\sin \angle BCH}=\frac{BC}{MB}\cdot\frac{\cos \angle BAC}{\cos \angle ABC}$
Since $AM=MB$, $\displaystyle \frac{\cos \angle BAC}{\cos \angle ABC}=\frac{AC}{BC}=\frac{\sin\angle ABC}{\sin\angle BAC}$.
Therefore, $\sin 2\angle BAC=\sin 2\angle ABC$. As the triangle is not isosceles, $2\angle BAC+2\angle ABC=180^\circ$. The result follows.
