A problem about a quadrilateral and diagonals in Kiselev's Geometry (Exercise 521).

Question

The problem is from Kiselev's Geometry exercise 521:

In a quadrilateral $ABCD$, through the midpoint of the diagonal $BD$, the line parallel to the diagonal $AC$ is drawn. Suppose that this line intersects the side $AB$ at a point $E$. Prove that the line $CE$ bisects the area of the quadrilateral.

[Edited] I originally thought the intersection should be on the extension of the side, but as @Aqua has pointed it out, it does not hold in that case.

Here is my geogoebra figure of the exercise. I found that depending on the location of the points, the intersection point might be on the side $AD$:

My attempt was to assign variables to the ratios between the segments divided by the diagonals, but it became way too cumbersome to calculate the ratio of the area. Another problem was to find which part of the quadrilateral will become a triangle by the line $CE$.

Any help would be much appreciated.

Answer 1

Solution to modified problem.

Let parallel to $AC$ through $D$ cuts line $AB$ at $F$.

Then $Area(ACD) = Area(FCA)$ (brown and green triangles, they have the same base $AC$ and high). Then $Area(AECD) = Area(FEC)$.

Now since $ME||DF$ we see that $E$ halves $BF$ which means that triangles $FEC$ and $EBC$ have equal area and we are done.

Answer 2

The statement does not hold. Move $A\to D$. If $A=D$ the quadrilateral is actually triangle $ABC$ and $CM$ halves the area of it and not $CE$.