It suffices to show that $CM \bot HN$. Click to reveal hidden text

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Using George_54 figure , easy to see that $AB$ is tangent to circumcircle of $\triangle{CKH}$ so let the this circle cut $CM$ again at $L'$ we have $ML'$ x $MC = MH^2 = MB^2$ so $\angle{MHL'} = \angle{HCL'}$ and $\angle{MBL'} = \angle{L'CB}$ thus we get $\angle{HL'B} = 180-w$ but since $\angle{KL'H} = w$ we must have $K,L',B$ collinear so $L' = L$ thus we get $\angle{CLH} = 90$ so $C,N,B,L$ is cyclic , so $\angle{ACB} = 2\angle{KCH} = 2\angle{KLH}= 2\angle{BCN} $

It is well know that there is a unique angle $\phi$ such that $\frac{sin\phi}{sin(\alpha-\phi)}=k$ for some constant $k$ and $\alpha$. Now note that $\triangle CHK$ is similar to $\triangle CBH$ so $\frac{BC}{CH}=\frac{BH}{KH}$ and when we apply sine theorems to $\triangle BCH$ and $\triangle BKH$ (and also the sine theorem for 2 angles built by median in $\triangle BCH$) we get that $\angle HKB=\angle HCM=\angle HCL$ so $CKHL$ is cyclic so $\angle CLH=90$ so $CLBN$ is cyclic. Now let $KH$ and $BN$ intersect at $X$. We have that $CKXB$ is cyclic so $\angle HXN=180-\angle ACB$. Also note that $HX$ and $BX$ are tangents to the circumcircle of $\triangle CHB$ so $CX$ is symmedian in that triangle so we get $ \angle HCX=\angle MCB=\angle XNH$ so $XHCN$ is also cyclic so $\angle HCN=180-\angle HXN=\angle ACB$ so we have $\angle BCN=\angle ACH$ so we are finished.

Obviously $\bigtriangleup HAK\sim \bigtriangleup CBH\rightarrow \frac{HA}{AK}=\frac{BC}{BH}\rightarrow \frac{BA}{AK}=\frac{BC}{BH}$ and $\angle BAK=\angle CBM\rightarrow \bigtriangleup BAK\sim \bigtriangleup CBM$. $\angle HKL=\angle AHK-\angle ABK=\angle BCH-\angle BCM=\angle HCL\rightarrow$ $HLCK$ is cyclic$\rightarrow \angle CLN=90^{\circ}\rightarrow CLBN$ is cyclic $\rightarrow \angle ACB=2\angle KCH=2\angle KLH=2\angle BLN=2\angle BCN.$