Nice and easy.

Let $M'$ be the point on $AB$, such that $M'L\perp AB$. By the angle bisector theorem on $\triangle ABC$, we have $$\frac{AC}{AB}=\frac{CL}{LB}\quad (1)$$Also $\triangle ABC\sim\triangle CHB\sim \triangle LM'C$, since $\angle ABC=\angle LM'C=\angle CHB=90^{\circ}$ and $\angle CBA=\angle M'BL=\angle HBC$. Therefore, we have $$\frac{AC}{AB}=\frac{CH}{CB}\quad (2)$$and $$\frac{CB}{LB}=\frac{HB}{M'B}\Longleftrightarrow \frac{LB+CL}{LB}=\frac{M'B+HM'}{M'B} \Longleftrightarrow\frac{CL}{LB}=\frac{HM'}{M'B} \quad (3)$$By $(1)$, $(2)$ and $(3)$, we conclude that $$\frac{CH}{CB}=\frac{HM'}{M'B},$$whence by the converse of the angle bisector theorem, we conclude that $M'\equiv M$. We have $$\angle (AL,MC)=180^{\circ}-\angle LAC-\angle ACM=180^{\circ}-\frac{\angle BAC}{2}-\angle ACK-\angle KCM=180^{\circ}-\frac{\angle BAC+ \angle ACB+\angle ABC}{2}=90^{\circ},$$since $\angle ACH=90^{\circ}- \angle HCB=90^{\circ}- (90^{\circ}- \angle CBA)=\angle CBA$ and $\angle HCM=\angle MCB$. Hence, $AK\perp CM$ and because we had $CH\perp AM$, we conclude that $K$ is the orthocentre of $\triangle ACM$ and furthermore, $MK\perp AC\perp BC\implies KM\parallel BC$ and since we had before that $CH\perp AB\perp LM\implies LM\parallel CH$, we conclude that $CKML$ is a parallelogram, which means that $CK=ML$.

Proposed by Alexey Doledenok

Nice and easy Let $T$ be the intersection of $KL$ and $CM$. We have that $x=\angle BAC=\angle HCB$, then we have that $\frac{1}{2}x=\angle LAB=\angle MAL = \angle MCL$, thus we have that $AMLC$ is cyclic. Thus we have that $\angle AML = 90$, this implies that $CH \parallel LM$. Since we have that $\frac{1}{2}x=\angle BAL=\angle HAT=\angle TCH$, thus we have that $ACTH$ is cyclic. This implies that $TH=TC$ and thus we have that $\angle THM = 90-\frac{1}{2}x$, but notice that $\angle CML=\angle TML = \frac{1}{2}x$, thus this implies that $\angle TMH = 90-\frac{1}{2}x$ this implies that $TH=TM$ this implies that $TM=TC$.Since we have that $CK \parallel LM$ this implies that we can take a homothety centered at $T$ with coefficient $-1$, thus we get that $CK=LM$.

$\angle ACM=90-\angle MCB=90-\angle MCH=\angle HMC=\angle AMC$ So $AC=AM$ from which $CL=ML$. Now $\angle CKL=\tfrac {\angle A}{2}+\angle ACH=90-\tfrac {\angle A}{2}=\angle CLK$. So we also have $CK=CL$ and therefore $CK=CL=ML$

with a stolen figure from djmathman

Metropolises 2020 P1 wrote: In a triangle $ABC$ with a right angle at $C$, the angle bisector $AL$ (where $L$ is on segment $BC$) intersects the altitude $CH$ at point $K$. The bisector of angle $BCH$ intersects segment $AB$ at point $M$. Prove that $CK=ML$ Note that $$\angle KCM=\angle HCB/2=\angle A /2=\angle HAK=90^\circ-\angle CKL$$hence $AL \perp CM$. Combining this with $\angle CAK=\angle KAM$ we obtain that $AK$ is the perpendicular bisector of $CM$. Therefore $\angle KMC=\angle KCM=\angle MCL$ implying that $KM \parallel CL \, (*)$. Similarly, $\angle LMC=\angle LCM=\angle KCM$ hence $KC \parallel LM, \, (**)$. Combining $(*)$ with $(**)$ we conclude that $CKML$ is a parallelogram, hence $KC=ML$ as desired.

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Quite easy problem.

Too easy. $\angle LCM=\angle LAM=\angle LAC=\angle LMC$, implies that $LM=LC$, and by easy angle chasing $CL=CK$, which completes the solution.