$ABCD$ is parallelogram. $\angle ABC=105^{\circ}$. $\angle CMD=135^\circ$. Find $\angle BKC$ 
$ABCD$ is parallelogram. $\angle ABC=105^{\circ}$. $BMC$ is equilateral triangle and $\angle CMD=135^\circ$. $K$ is midpoint of $AB$. Find $\angle BKC$

My attempts

1) $\angle MBC=\angle MCB=\angle BMC=60^{\circ}$
2) $\angle MCD=15^{\circ}$, $\angle MDB=30^{\circ}$
I want to prove that $MB \perp CK$ but I need help here

Let $N$ be the point on $CD$ such that $\angle CMN=15^\circ$.
Then $\angle CMN=15^\circ=\angle MCN$ and $\angle MDN=30^\circ=\angle MND$.
So $DM=MN=NC$.
As $BC=BM$, $MN=CN$ and $BN=BN$, $\triangle BCN\cong \triangle BMN$ and therefore, $\angle CBN=\angle MBN=30^\circ$.
So, $\angle BNC=75^\circ=\angle BCN$ and hence $BC=BN=BM$.
As $\angle ADM=75^\circ=\angle BCN$, $BC=AD$ and $DM=NC$, $\triangle BCN\cong \triangle ADM$. 
So, $AM=BN=BM$.
Therefore, $\angle BAM=\angle ABM=45^\circ$.
Since $K$ is the midpoint of $AB$, $MK\perp AB$.
$\angle KMB=180^\circ-90^\circ-45^\circ=45^\circ=\angle KBM$.
Therefore, $BK=KM$.
$\triangle KBC\cong\triangle KMC$.
$\angle BKC=\angle MKC=90^\circ\div 2=45^\circ$.

