CEMC Fermat, 1999

Geometry Difficulty 3.0 AMC 10/12 Find the answer Canada

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

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

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