Let $A'$ be the $A-$antipode; then $K = A'M \cap (ABC)$, and if $X' \in A'M$ such that $\angle{A'CX'} = \angle{BCA}$, then $\triangle{MCX'} \sim \triangle{MKX} \implies \frac{MX'}{X'A'} = \frac{MX}{XA} \implies$ it suffices to show that $\frac{MX'}{X'A'} = \frac{MY}{YB}$. By the Ratio Lemma on $\triangle{MCA'}, \triangle{MAB}$, this is equivalent to $\frac{AM}{AB} \cdot \frac{\cos \angle{C}}{\cos \angle{B}} = \frac{CM}{CA'} \cdot \frac{\cos \angle{C}}{\sin \angle{C}}$. Since $AM = CM$, this simplifies to $AB \cos \angle{B} = A'C \sin \angle{C}$, but this is obvious since $\frac{AB}{A'C} = \frac{\sin \angle{C}}{\cos \angle{B}}$ by angle chasing in $(ABC)$. $\square$

Let $AY\cap BX = T$ Note that $A,K,M,T$ are cyclic Hence $\angle ATM=90^{\circ}$ So $MT$ passing through the mid-point of $AB$ Which means that $\dfrac{MX}{XA}\cdot \dfrac{BY}{YM}=1$ So $XY//AB$

SQTHUSH wrote: Let $AY\cap BX = T$ Note that $A,K,M,T$ are cyclic Hence $\angle ATM=90^{\circ}$ So $MT$ passing through the mid-point of $AB$ Which means that $\dfrac{MX}{XA}\cdot \dfrac{BY}{YM}=1$ So $XY//AB$ How to prove that $A,K,M,T$ are cyclic?

dictator2002 wrote: SQTHUSH wrote: Let $AY\cap BX = T$ Note that $A,K,M,T$ are cyclic Hence $\angle ATM=90^{\circ}$ So $MT$ passing through the mid-point of $AB$ Which means that $\dfrac{MX}{XA}\cdot \dfrac{BY}{YM}=1$ So $XY//AB$ How to prove that $A,K,M,T$ are cyclic? $\angle BKM=90^{\circ}-\angle ACB=\angle CAT$

SQTHUSH wrote: dictator2002 wrote: SQTHUSH wrote: Let $AY\cap BX = T$ Note that $A,K,M,T$ are cyclic Hence $\angle ATM=90^{\circ}$ So $MT$ passing through the mid-point of $AB$ Which means that $\dfrac{MX}{XA}\cdot \dfrac{BY}{YM}=1$ So $XY//AB$ How to prove that $A,K,M,T$ are cyclic? $\angle BKM=90^{\circ}-\angle ACB=\angle CAT$ Thanks