67. As shown in the figure, in isosceles $\triangle A B C$, $\angle A C B=90^{\circ}, M, N$ are two points on the hypotenuse $A B$, and $\angle M C N=45^{\circ}$. Given $A M=3, B N=5$, then $M N=$ $\qquad$ -
Answer: $\sqrt{34}$

Solution: As shown in the figure, rotate $\triangle B N C$ 90 degrees clockwise around point $C$
$$
\angle C A N^{\prime}=\angle B, A N^{\prime}=B N=5,
$$

Therefore, $\angle M A N^{\prime}=90^{\circ}, M N^{\prime}=\sqrt{A M^{2}+A N^{\prime 2}}=\sqrt{34}$.

Also, since
$$
\angle M C N=45^{\circ} \text {, }
$$

we have
$$
\angle M C N^{\prime}=\angle M C A+\angle A C N^{\prime}=45^{\circ},
$$

which means
$$
\angle M C N=\angle M C N^{\prime} \text {, }
$$

Therefore,
$$
\triangle M C N \cong \triangle M C N^{\prime},
$$

Thus,
$$
M N=M N^{\prime}=\sqrt{34} \text { . }
$$