It is given that in triangle $A B C$, $\angle B A C=90^{\circ}, A D$ is the altitude from $A$ on to $B C$.
Since, $A B=15$ and $B C=25$
$\therefore \quad A C=\sqrt{B C^2-A B^2}=\sqrt{625-225}$
$=\sqrt{400}=20$
Now, since area of $\triangle A B C=\frac{1}{2}(B C)(A D)$
$=\frac{1}{2}(A B)(A C)$
$\Rightarrow \frac{1}{2}(B C)(A D)=\frac{1}{2} \times 15 \times 20$
$\Rightarrow \quad 25 \times A D=300$
$\Rightarrow \quad A D=12$
$\because A E D F$ is a rectangle, then
$E F=A D=12$
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