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.

Let $O$ be the circumcenter of $ABC$. We will use the following fact:

The circumcenter $O$ of a triangle $ABC$ is isogonally conjugate to its orthocentre $S$. That is, if $I$ is the incenter of $ABC$, we have $\measuredangle IAO=\measuredangle SAI$, $\measuredangle IBO=\measuredangle SBI$ and $\measuredangle ICO=\measuredangle SCI$.

In this particular instance, this means that from $\measuredangle HCL=\measuredangle MCL$, we can infer that $O$ must lie on $CM$. Since it must also lie on the perpendicular bisector of $\overline{AB}$, it follows that $O=M$. From there, we can conclude that $\measuredangle ACB=90^{\circ}$ by the converse of Thales' Theorem.

The reverse direction of $\measuredangle ACB=90^{\circ}\Rightarrow \measuredangle HCL=\measuredangle MCL$ is just as easy.

Proof of the fact: Let $I$ be the incenter of $ABC$. We will prove that $\measuredangle IAO=\measuredangle SAI$. The proof for the other corners of $ABC$ is analogous. First, since $AI$ bisects $\measuredangle BAC=:\alpha$ and $\measuredangle OAC=90^{\circ}-\frac{\measuredangle COA}{2}=90^{\circ}-\measuredangle CBA=:90^{\circ}-\beta$, we have $\measuredangle IAO=\frac{\alpha}{2}-90^{\circ}+\beta$. Now let $D$ be the foot of the altitude from $A$ to $BC$. Then we similarly have $\measuredangle BAS=\measuredangle BAD=90^{\circ}-\measuredangle DBA=90^{\circ}-\beta$ and thus $\measuredangle SAI=\frac{\alpha}{2}-90^{\circ}+\beta$ so that $\measuredangle IAO=\measuredangle SAI$, as claimed.

Let's assume $\angle C > 90 ^\circ$ as show in the picture. $CL1$ is the bisector of $\angle ACB$ , $CL2$ is the bisector of $\angle HCM$ , now we show $AL2>AL1$

let $l1=AL1,l2=AL2$

$l1=\frac{AC}{AC+BC}*(c1+c2)$, $AC=\sqrt{h^2+c1^2},BC=\sqrt{h^2+c2^2}$, $l1=\frac{\sqrt{h^2+c1^2}}{\sqrt{h^2+c1^2}+\sqrt{h^2+c2^2}}*(c1+c2)=\frac{\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}-(h^2+c1^2)}{c2^2-c1^2}*(c1+c2)=\frac{\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}-(h^2+c1^2)}{c2-c1}$

$HM=\frac{c2-c1}{2},m=\sqrt{h^2+(\frac{c2-c1}{2})^2}$

$AL2=c1+HL2=c1+\frac{h}{h+m}*(\frac{c2-c1}{2})=c1+\frac{h}{h+\sqrt{h^2+(\frac{c2-c1}{2})^2}}*(\frac{c2-c1}{2})=c1+\frac{h*(\sqrt{h^2+(\frac{c2-c1}{2})^2}-h)}{(\frac{c2-c1}{2})^2}*(\frac{c2-c1}{2})=c1+\frac{h*(\sqrt{4h^2+(c2-c1)^2}-2h)}{c2-c1}=\frac{c2c1-c1^2+h*(\sqrt{4h^2+(c2-c1)^2}-2h)}{c2-c1}$

$AL2-AL1=\frac{D}{c2-c1}$ $D=c2c1-c1^2+h*(\sqrt{4h^2+(c2-c1)^2}-2h)-(\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}-(h^2+c1^2))=c1c2-h^2+h*\sqrt{4h^2+(c2-c1)^2}-\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}=c1c2-h^2+\frac{3h^2(h^2-c1c2)}{h*\sqrt{4h^2+(c2-c1)^2}+\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}}=(c1c2-h^2)(1-\frac{3h^2}{h*\sqrt{4h^2+(c2-c1)^2}+\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}})$

I left last step , note $h^2<c1c2$, when it is right angle $h^2=c1c2$