ABCD is a rectangle and inside it a circle is inscribed touching its larger sides, AD and BC. BD is the rectangle's diagonal which intersects the circle in points E and F, forming the chord EF.
Knowing that BE = 3, EF = 9 and FD = 16, find the side BC. I'm currently trying to solve this by finding the radius, knowing that it's a simple matter of applying the Pythagorean theorem. But I'm struggling in expressing the information I have in an helpful way, maybe expressing the diagonal and circle as functions could help? I've been spending some time on this with no avail.
It would look something like this:
Let $|AD|=|BC|=a$, $|AB|=|CD|=b$, $|BG|=|AH|=u$, $|BE|=p=3$, $|EF|=h=9$, $|FD|=q=16$, $|BD|=d=p+h+q=28$.
By the power of a point $B$,
\begin{align} u&=\sqrt{p(p+h)}=6 . \end{align}
By the power of a point $D$,
\begin{align} a&=u+\sqrt{q(q+h)}=26 ,\\ b&=\sqrt{d^2-a^2}=6\sqrt3 . \end{align}