59. In the square $\mathrm{ABCD}$, point $\mathrm{E}$ is on $\mathrm{BC}$, $B E=2, C E=1$, point $P$ moves on $\mathrm{BD}$, then the minimum value of $\mathrm{PE}+\mathrm{PC}$ is $\qquad$.
Answer: $\sqrt{13}$
Solution: Take a point $E^{\prime}$ on $\mathrm{AB}$ such that $B E^{\prime}=B E$, then $P E=P E^{\prime}$,
When $P, C, E^{\prime}$ are not collinear,
$$
P E^{\prime}+P C>C E^{\prime} ;
$$

When $E^{\prime}, P, C$ are collinear, $P E+P C$ is minimized, and $P E+P C=P E^{\prime}+P C=C E^{\prime}$,
Since $B E^{\prime} \perp C B$, we have $(P E+P C)^{2}=B E^{\prime 2}+B C^{2}$,

That is
$$
(P E+P C)^{2}=2^{2}+3^{2}=13, P E+P C=\sqrt{13},
$$
$P E+P C$ has a minimum value of $\sqrt{13}$.