73. As shown in the figure, in quadrilateral $A B C D$, diagonal $A C \perp B D$, the lengths of the four sides are $A B=6, B C=m$, $C D=8, D A=n$, then $m^{2}+n^{2}=$ $\qquad$
Answer: 100.
Solution: Let the intersection of the diagonals be $O$. According to the Pythagorean theorem, we have
$$
\begin{array}{l}
m^{2}+n^{2} \\
=\left(O B^{2}+O C^{2}\right)+\left(O D^{2}+O A^{2}\right) \\
=\left(O A^{2}+O B^{2}\right)+\left(O C^{2}+O D^{2}\right) \\
=6^{2}+8^{2} \\
=100
\end{array}
$$