5. As shown in Figure 1, in $\triangle A B C$, $\angle B=$ $45^{\circ}, A C$'s perpendicular bisector intersects $A C$ at point $D$, and intersects $B C$ at point $E$, and $\angle E A B: \angle C A E=3: 1$. Then $\angle C$ equals ( ).
(A) $27^{\circ}$
(B) $25^{\circ}$
(C) $22.5^{\circ}$
(D) $20^{\circ}$
5.A.

Since $E D \perp$ bisects $A C$, therefore,
$$
\begin{array}{l}
\angle C=\angle E A C, \\
\angle C+\angle C A B=180^{\circ}-45^{\circ}=135^{\circ} .
\end{array}
$$

And $\angle C A B=\angle C A E+\angle E A B=4 \angle C A E=4 \angle C$, then $\angle C+\angle C A B=\angle C+4 \angle C=5 \angle C=135^{\circ}$. Thus, we get $\angle C=\frac{135^{\circ}}{5}=27^{\circ}$.