G38 (18-1, Okslovakia) In a planar convex quadrilateral with an area of 32 square centimeters, the sum of the lengths of two opposite sides and one diagonal is 16 centimeters. Determine all possible lengths of the other diagonal.

保留源文本的换行和格式，翻译结果如下：

G38 (18-1, Okslovakia)  
In a planar convex quadrilateral with an area of 32 square centimeters, the sum of the lengths of two opposite sides and one diagonal is 16 centimeters. Determine all possible lengths of the other diagonal.
Solve: In a convex quadrilateral $ABCD$, let $AB + AC + CD = 16$. We need to find $BD$.
Given that $32=$ the area of quadrilateral $ABCD$
$$
\begin{array}{l}
=S_{\triangle ABC} + S_{\triangle ACD} \\
=\frac{1}{2} AB \cdot AC \cdot \sin \angle BAC + \frac{1}{2} AC \cdot CD \cdot \sin \angle ACD \\
\leqslant \frac{1}{2} AC(AB + CD) \\
\leqslant \frac{1}{8}(AC + AB + CD)^2 \\
=\frac{256}{8} = 32.
\end{array}
$$

Thus, the equality holds, which requires $\angle BAC = \angle ACD = 90^\circ$, and $AC = AB + CD = 8$.

Construct a rectangle $AB A' C$ outward with $AC$ and $AB$ as sides. It is easy to see that $A'B = AC = 8$, $DA' = DC + AB = 8$, and $\angle DA'B = 90^\circ$. Therefore, in the isosceles right triangle $BA'D$, $BD = 8\sqrt{2}$ cm.