It is known that in the triangle $ABC$ the smallest side is $BC$. Let $X, Y, K$ and $L$ - points on the sides $AB, AC$ and on the rays $CB, BC$, respectively, are such that $BX = BK = BC =CY =CL$. The line $KX$ intersects the line $LY$ at the point $M$. Prove that the intersection point of the medians $\vartriangle KLM$ coincides with the center of the inscribed circle $\vartriangle ABC$.