It is given that in triangle $A B C$, $\angle BAC=90°,AD$ is the altitude from A on to B C.

since, $A B=15$ and $B C=25$$\therefore AC=\sqrt{B{C}^2-A{B}^2}=\sqrt{625-225}$

Now, since area of $∆ABC=\frac{1}{2}\left(BC\right)\left(AD\right)$

$=\frac{1}{2}\left(AB\right)\left(AC\right)$

$\Rightarrow \frac{1}{2}\left(BC\right)\left(AD\right)=\frac{1}{2}\times 15\times 20$

$\Rightarrow 25\times AD=300$

$\Rightarrow AD=12$

$\because AEDF$ is a rectangle, then $E F=A D=12$