Problem 1485

Let $ABC$ be an acute-angled triangle. The feet of the altitudes from $A$, $B$ and $C$ are $D$, $E$ and $F$ respectively. Prove that $DE + DF \leq BC$ and determine the triangles for which equality holds.

The altitude from $A$ is the line through $A$ which is perpendicular to $BC$. The foot of this altitude is the point $D$ where it meets $BC$. The other altitudes are similarly defined.