Apply Barycentric w.r.t. $\triangle ABC$. We have $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. For all their areas to be equal, we must have $M=(\tfrac 13,\tfrac 13,\tfrac 13)$. Now, let the midpoints of $AB,BC,CA$ be $D=(\tfrac 12,\tfrac 12,0),E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 12,0,\tfrac 12)$ respectively. We then proceed by finding the equations of $AD,BE,CF$. By Barycentric Ceva, they meet at a point $P=(\tfrac 13,\tfrac 13,\tfrac 13)$. Since $M=P$ we are done.