18.11 Given a quadrilateral $A B C D$ with a point $E$ inside it. Connecting $A E, B E, C E, D E$ divides the quadrilateral $A B C D$ into four triangles of equal area. Then, among the statements:
甲: $A B C D$ is a convex quadrilateral,
乙: $E$ is the midpoint of diagonal $A C$ or diagonal $B D$,
丙: $A B C D$ is a parallelogram,

which of the following is true?
(A) Only 甲 is correct.
(B) Only 乙 is correct.
(C) 甲, 乙, and 丙 are all correct.
(D) 甲, 乙, and 丙 are all incorrect.
(China Junior High School Mathematics League, 1987)
[Solution 1] If $ABCD$ is a concave quadrilateral with $AC$ as the axis of symmetry, it is easy to see that the midpoint of $AC$ has the property required for point $E$ in the problem. Therefore, neither A nor C is correct.

Assume that $AE, BE, CE, DE$ divide quadrilateral $ABCD$ into four triangles of equal area.
From $S_{\triangle ABE}=S_{\triangle ADE}$, we get that the distances from $B$ and $D$ to line $AE$ are equal.
Let the intersection of line $BD$ and $AE$ be $F$, then $F$ is the midpoint of $BD$, i.e., $E$ lies on line $AF$.
If $F$ coincides with $E$, then $E$ is the midpoint of $BD$, and B is true.
If $F$ does not coincide with $E$, similarly, from $S_{\triangle BEC}=S_{\triangle DEC}$, we get that $E$ lies on line $CF$, meaning that $A, C$ are both on line $EF$, so $A, E, F, C$ are collinear.
Again, from $S_{\triangle ABE}=S_{\triangle BEC}$, we get $AE=EC$. So $E$ is the midpoint of $AC$, and B is true.
Therefore, the answer is $(B)$.
[Solution 2] As shown in the figure on the next page, point $E$ is inside quadrilateral $ABCD$, and line segment $AE$ is also inside the figure, i.e., ray $AE$ is within $\angle BAD$.
Thus, points $B$ and $D$ are on opposite sides of ray $AE$.
From $S_{\triangle ABE}=S_{\triangle ADE}$, we know that the distances from $B$ and $D$ to $AE$ are equal, and the intersection of $AE$ and $BD$, $F$, is the midpoint of $BD$.

If $F$ coincides with $E$, then $E$ is also the midpoint of $BD$. Then B is true.

If $F$ does not coincide with $E$, similarly, from $S_{\triangle BEC}=S_{\triangle DEC}$, we get that $CE$ also passes through the midpoint $F$ of $BD$, i.e., $A, C$ are both on line $EF$, so $A, C, E, F$ are collinear.
Thus, from $S_{\triangle AEB}=S_{\triangle BEC}$, we know that point $E$ is the midpoint of $AC$, and B is true. Therefore, the answer is $(B)$.