1999 Fermat Problem 17

In \triangle ABC, AC = AB = 25 and BC = 40. D is a point chosen on BC. From D, perpendiculars are drawn to meet AC at E and AB at F. DE + DF equals

\textbf{(A)}\ 12\quad \textbf{(B)}\ 35\quad \textbf{(C)}\ 24\quad \textbf{(D)}\ 25\quad \textbf{(E)}\ \frac{35}{2}\sqrt{2}

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