Square $ABCD$ has side lengths of 13 units. Point $E$ lies in the interior of the square such that $AE = 5$ units and $BE = 12$ units. What is the distance from $E$ to side $AD$?
Since $5^2+12^2=13^2$, triangle $AEB$ is a right triangle.  Define $F$ to be the foot of the perpendicular drawn from $E$ to side $AB$.  The distance from $E$ to side $AD$ is $AF$.  By the similarity of triangles $AEF$ and $ABE$, \[
\frac{AF}{AE}=\frac{AE}{AB}.
\]Solving for $AF$, we find $AF=AE^2/AB=5^2/13=\boxed{\frac{25}{13}} = \boxed{1\frac{12}{13}}$ units.

[asy]
unitsize(1.5mm);
defaultpen(linewidth(.7pt)+fontsize(10pt));
dotfactor=3;
pair A=(0,0), B=(13,0), C=(13,13), D=(0,13), E=(1+12/13,sqrt(5^2-(1+12/13)^2)), F=(1+12/13,0);
pair[] dots={A,B,C,D,E,F};
draw(A--B--C--D--cycle);
draw(A--E--B);
draw(E--F);
dot(dots);
label("A",A,SW);
label("B",B,SE);
label("C",C,NE);
label("D",D,NW);
label("E",E,N);
label("F",F,S);
[/asy]