ismayilzadei1387 wrote: what is definition for $C'$? Edited. Thank you.

Clearly $M$ is center of $(AB'B)$ therefore $AM=MB'$, now by Reim's theorem $A,K,L$ are colinear and now notice that $\angle MLC=\angle MB'A=\angle BAC$ which implies that $AMLC$ is a paralelogram because $AM \parallel LC$ is already true, now because $AM=MB$ we have that $BMCL$ is a paralelogram as well and now $-1=(A, B; M, \infty_{AB}) \overset{L}{=} (KL \cap CM, \infty_{CM}; M, C)$ therefore $KL$ bisects $CM$ as desired thus we are done .

Claim: $L,K,A$ are collinear. Proof: \[\angle LkD=\angle LCD=\angle LCB=\angle B=\angle ABD=180-\angle DKA\] Claim: $NM^2=NK.NA$ Proof: \[\angle KMN=\angle KLC=\angle ALC=\angle NAM\] Claim: $NC^2=NK.NA$ Proof: $\angle NCK=\angle MCK=\angle MB'K$ $\angle MB'K+\angle NAM=\angle MB'K+\angle KMC=\angle MB'K+\angle KB'A=\angle MB'A=\angle B'AM= \angle B'AN+\angle NAM$ Thus $\angle NCK=\angle MB'K=\angle B'AN$

Claim: $L$ is the reflection of $A$ about the midpoint of $AC$ Let $N$ be the midpoint of segment $MC$. Then let $A'$ be the reflection of $A$ about $N$. Then since, $$\angle MA'C=\angle MAC=\angle MAB'=\angle AB'M=180^{\circ}-\angle MB'C$$the point $M$, $B'$, $C$, and $A'$ are concyclic. Since $CA'||AB$ and $A'$ lies on $(MB'C)$ we conclude that $A'=L$. Claim: $K$ is the intersection point of $AL$ with $(MB'C)$ Let $K'$ be the intersection point of $AL$ with $(MB'C)$. Then since, $$\angle ABD=\angle ABC=\angle BCL=\angle DCL=\angle DK'L=180^{\circ}-\angle DK'A$$the points $B$, $D$, $K'$, and $A$ lie on a circle. Since $K'$ also lies on $(MB'C)$ we can conclude that $K'=K$. The result follows.

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Solution

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By Reim's Theorem, points $A$, $K$ and $L$ are collinear since $$\angle AKD=\pi-\angle ABC=\pi-\angle BCL=\pi-\angle DKL$$On the other hand, since $MDCB'$ is cyclic $$\angle MLC=\angle MDC=\angle MB'A=\angle BAC$$Thus, $AMLC$ is a parallelogram which implies $MB=MA=CL$. Since it is parallelogram, $KL$ cuts $MC$ half, as desired.

Sol