<s>[INST] 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.

 [/INST] 

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  $h^2=c1c2$ 
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