Four, (14 points) As shown in the figure, the angle bisector $B E$ of $\triangle A B C$ and the median $A D$ of side $B C$ are perpendicular and equal. Given $B E = A D = 4$. Find the three sides of $\triangle A B C$.
Since $B E$ is the bisector of $\angle B$, and $A D \perp B E$, it is known that points $A$ and $D$ are symmetric about line $B E$.

Taking $B E$ as the axis of symmetry, construct the symmetric point $F$ of point $C$ about $B E$. Clearly, line segment $D F$ is symmetric to $A C$, and their intersection point is $\pm$. Point $H$ is the intersection of the extension of $B E$ with $F C$, as shown in the right figure.

Clearly, $\triangle B F C$ is an isosceles triangle, $A D$ is its midline, $B H$ bisects and is perpendicular to $F C$, so $E$ is the centroid of $\triangle B F C$. Therefore, $A E=\frac{1}{2} E C, E H=\frac{1}{2} B E=2$, and
$$
H C=\frac{1}{2} F C=A D=4 . A B=\frac{1}{2} B C .
$$

In Rt $\triangle B H C$,
$$
B C=\sqrt{B H^{2}+H C^{2}}=2 \sqrt{13} \text {, }
$$

Thus, $A B=\frac{1}{2} B C=\sqrt{13}$.
In Rt $\triangle E H C$,
$$
E C=\sqrt{E H^{2}+H C^{2}}=2 \sqrt{5} \text {, }
$$

Therefore, $A C=A E+E C=3 \sqrt{5}$.
Note: There are many ways to prove this, for example:
(1) Using area proof (since $\triangle A B E=\triangle B E D=$ $\triangle D E C=\frac{1}{2} A O \cdot B E=4$, thus $\triangle A B C=12$, $\triangle A B D=6$. Then find the three sides),
(2) Using segment length formulas (by the midline length formula and the angle bisector length formula, set up a system of equations to solve),
(3) Using the midline (as shown in the right figure, construct $D K / / B E$, since $D K$ is the midline of $\triangle B E C$, and $O E$ is the midline of $\triangle A D K$, thus solve),
(4) Using Menelaus' theorem ($B E$ with respect to $\triangle A D C$, $A D$ with respect to $\triangle B C E$),
(5) Applying the cosine rule.
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It is evident that there are many proofs for this, and readers are encouraged to try them, which will be beneficial.