If ABC is a self-conjugate triangle. D, E, F are the mid-points of AC, CB, AB respectively. Does CF cut DE into 2-halves?
If CF meets DE at G then the triangles CDG, CAF are similar, as are the triangles CEG, CBF. In each case, the sides of the smaller triangle are half as long as those of the larger triangle. Therefore DG=½AF=½BF=EG.