In triangle ABC the point D is the midpoint of AB and the point E is the midpoint of AC. F is the midpoint of DE, and the line AF is produced to meet BC at G. What is true of the point G?

By the mid-point theorem DE is parallel to $BC$, so triangle ADE is the half-size copy of triangle $ABC$ sitting at the vertex A. $F$ is the midpoint of DE, which corresponds to the midpoint $M$ of $BC$ under that copying, and corresponding points lie on a common line through A. Hence A, $F$ and $M$ are collinear and G is exactly $M$, the midpoint of $BC$ — in other words $AF$ lies along the median from A. (Vectors say the same thing: $F=(2A+B+C)/4$, so $F-A$ = ½$(M-A)$ where $M=(B+C)/2$.) The altitude foot would need extra information about the angles, and neither 1 : 2 division works because the halving happens along the cevian, not along $BC$.