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$ABCD$ 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 Figure

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

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    $\begingroup$ I think you mean $\angle MDC = 30^{\circ}$ $\endgroup$ – idok Jan 12 '18 at 10:18
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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, $BK=KM$.

$\triangle KBC\cong\triangle KMC$.

$\angle BKC=\angle MKC=90^\circ\div 2=45^\circ$.

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WLOG, I assumed $|BK| = |AK| = 1$ (We could say $a$ or $x$ or whatever we want to but we don't lose generality and this way makes things more clear I believe). Then if we apply The Law of Sines in $\Delta CDM$, we have: $$\frac{|CD|}{\sin135^{\circ}} = \frac{|CM|}{\sin30^{\circ}} \implies 2\sqrt{2} = 2|CM| \implies |CM| = \sqrt{2}$$ Now, notice that $\angle MBK = 45^\circ$ so $|KM|$ is nothing but $1$ (We can see that by using Law of Cosines or noticing that a triangle like that one must be unique). Then $\Delta BKM$ is isosceles right angle triangle as you foresaw. So the result follows.

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