Hint: Construct the intersection point of the perpendicular bisector of $AB$ and $BM$ and find equal triangles.

Let P and Q be midpoints of KM and BC. Then <CPB=90 ( because CK=CM), and QP=QB=QC ( because <CPB=90). Let <ABM=x, <CBM=2x, <QMP=<MBA=x (QM||AB), <QPB=<QBP=2x (QP=QB), and <PQM=<QPB-<QMP=2x-x=x=<PMQ, thus triangle QPM is isosceles and QP=PM. BC=2QP=2PM=KM.

No_Name1 wrote: Let $P$ and $Q$ be midpoints of $KM$ and $BC$ respectively.Then $<CPB=90$ ( because $CK=CM$), and $QP=QB=QC$ ( because $<CPB=90$). Let $<ABM=x, <CBM=2x, <QMP=<MBA=x (QM||AB), <QPB=<QBP=2x (QP=QB)$, and $ <PQM=<QPB-<QMP=2x-x=x=<PMQ,$ thus triangle $QPM$ is isosceles and $QP=PM.$ $BC=2QP=2PM=KM.$ Nice proof! I really liked how you made midpoints, used parallel lines and iscolceles triangles to write everything in terms of each other, I never actually thought of that. I just didn’t understand how you got $QP=QC$ (or $QP=QB$,since $QC$ and $QB$ are equal) Can you explain that a bit more?

Yeetopedia wrote: No_Name1 wrote: Let $P$ and $Q$ be midpoints of $KM$ and $BC$ respectively.Then $<CPB=90$ ( because $CK=CM$), and $QP=QB=QC$ ( because $<CPB=90$). Let $<ABM=x, <CBM=2x, <QMP=<MBA=x (QM||AB), <QPB=<QBP=2x (QP=QB)$, and $ <PQM=<QPB-<QMP=2x-x=x=<PMQ,$ thus triangle $QPM$ is isosceles and $QP=PM.$ $BC=2QP=2PM=KM.$ Nice proof! I really liked how you made midpoints, used parallel lines and iscolceles triangles to write everything in terms of each other, I never actually thought of that. I just didn’t understand how you got $QP=QC$ (or $QP=QB$,since $QC$ and $QB$ are equal) Can you explain that a bit more? QP=QC as a median in right angled triangle CPB

Solution that I made during the contest. Let $T$ be a point on segment $CK$ such that $BT$ is angle bisector of $\angle CBK$. Let $\angle CBT=\angle TBK =\angle MBA = \alpha$ and $\angle BCT=\beta$ $\implies$ $\angle BTK=\alpha +\beta$ $\implies$ $\angle CKM=\angle CMK= 2\alpha+\beta$ $\implies$ $\angle BAC= \alpha+\beta$ $\implies$ $\triangle BKT\sim \triangle BMA$ $\frac{BK}{KT}=\frac{BM}{MA}=\frac{BK+KM}{CK}$ $\implies$ $\frac{CK}{KT}=\frac{BK+KM}{BK}$ and after minus $1$ from both equation fractions we get $\frac{CT}{KT}=\frac{KM}{KB}$ from angle bisector we know $\frac{CT}{KT}=\frac{BC}{BK}$ $\implies$ $\frac{KM}{BK}=\frac{BC}{BK}$ $\implies$ $BC=MK$

Let $C’$ be the image of reflection of $C$ across $BM$. Then $C’KCM$ becomes a rhombus and, consequently, $AC’KM$ is a parallelogram. Hence, $\angle C’AB=\angle ABM=\angle C’BA$ , i.e. $MK=C’A=C’B=BC$.

Trig Bash solution. Let $\angle ABM= \alpha$ and $\angle BMC = \beta$, so $\angle CBM = 2\alpha$ and $\angle CKM = \beta$. By the Law of Sines in $\triangle BCM$ and $\triangle BMA$ we have that $$1=\frac{CM}{MB} \cdot \frac{BM}{BA} = \frac{\sin 2\alpha}{\sin (2\alpha + \beta)} \cdot \frac{\sin (\beta - \alpha)}{\sin \alpha} \Leftrightarrow \sin (2\alpha + \beta ) = 2\cos \alpha \sin (\beta - \alpha ) = \sin \beta - \sin (2\alpha - \beta ) \Longrightarrow \frac{\sin \beta}{2\sin 2\alpha \sin \beta} = 1$$Now, by the Law of Sines in $\triangle BCK$ and $\triangle KCM$ we obtain that $$\frac{BC}{KM} = \frac{BC}{CK} \cdot \frac{CK}{KM} = \frac{\sin \beta}{\sin 2\alpha} \cdot \frac{\sin \beta}{\sin 2\beta} = \frac{\sin \beta}{2\sin 2\alpha \sin \beta} = 1$$that is, $BC=MK$. $\blacksquare$

Let $T$ be a point on $BM$ s.t. $AT=TB$.From angle chasing we get $\triangle ATM \sim \triangle CBK$,$AM=KC$ so $AT=BC$ and $AT=BT \implies BT=BC$,$TM=BK \implies BT=KM$ so we're done.