I don't believe there's enough information. Let's let
$$A=(0,0)$$
$$B=(16,0)$$
$$C=(9\cos\theta,9\sin\theta)\text{ for some }0<\theta<90^{\circ}$$
$$D=B+C=(16+9\cos\theta,9\sin\theta)$$
$$F=\left(\sqrt{15^2-9^2\sin^2\theta},9\sin\theta\right)$$
Then we have a parallelogram $ABDC$ with a point $F$ on $CD$, such that $|AC|=9$, $|AF|=15$, and $|AB|=16$. Our next step is to find $E$. We'll be done if we can show that $|AE|$ is a non-constant function of $\theta$.
$E$ is the intersection of the lines determined by segments $AF$ and $BD$. To find the coordinates of $E$, we'll first find the equations of these lines. Using the point-slope form, we have that the equation for the line determined by segment $AF$ is:
$$y=\frac{9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}}\cdot x$$
Using the point-slope form, we have that the equation for the line determined by segment $BD$ is:
$$y=\frac{9\sin\theta}{9\cos\theta}\cdot (x-16)$$
Hence we can find the $x$-coordinate of $E$ by solving
$$\frac{9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}}\cdot x=\frac{9\sin\theta}{9\cos\theta}\cdot (x-16)$$
This gives us that
$$x=\frac{16\cdot\sqrt{15^2-9^2\sin^2\theta}}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
We can then plug this into the equation for the line determined by segment $AF$ to obtain that
$$y=\frac{16\cdot9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
Hence
$$E=\left(\frac{16\cdot\sqrt{15^2-9^2\sin^2\theta}}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta},\frac{16\cdot9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}\right)$$
It follows that
$$|AE|=\frac{16\cdot15}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
Note that if $\theta=30^{\circ}$, then $|AE|\approx36.8$, but if $\theta=60^{\circ}$, then $|AE|\approx28.9$. So $|AE|$ is a non-constant function of $\theta$.
Finally, note that $m=|AE|-15$. So $m$ is a non-constant function of $\theta$. We need more information.