5. Given a quadrilateral $A B C D$ with a point $E$ inside it, connecting $A E, B E, C E$, and $D E$ divides the quadrilateral $A B C D$ into four triangles of equal area. Then, the proposition:
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.
Among the above three,
(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.
Answer ( )
5. (B)

Solution As shown in the figure, point $E$ is inside quadrilateral $ABCD$, and line segment $AE$ is also in the figure, meaning ray $AE$ is within $\angle BAD$.

Thus, points $B$ and $D$ are on opposite sides of ray $AE$.

Given $S_{\triangle ABE} = S_{\triangle ADE}$, it follows that the distances from $B$ and $D$ to $AE$ are equal. The intersection point $F$ of $AE$ and $BD$ is the midpoint of $BD$.

If $F$ coincides with $E$, then $E$ is also the midpoint of $BD$, and thus statement (ii) holds.

If $F$ does not coincide with $E$, similarly, from $S_{\triangle BEC} = S_{\triangle DEC}$, it follows that $CE$ also passes through the midpoint $F$ of $BD$, and points $A$ and $C$ are on line $EF$, meaning $A, C, E, F$ are collinear.

Therefore, from $S_{\triangle AEB} = S_{\triangle BEC}$, it follows that point $E$ is the midpoint of $AC$, and thus statement (ii) holds.