Trapezoid $ABCD$ is divided into a triangle and a parallelogram by segment $CE$, see the figure. Point $F$ is the midpoint of segment $CE$, line $DF$ passes through the midpoint of segment $BE$, and the area of triangle $CDE$ is $3 \, \text{cm}^2$.

Determine the area of trapezoid $ABCD$.

(E. Semerádová)

![](https://cdn.mathpix.com/cropped/2024_04_17_b1dd73eb0c32fed574d9g-4.jpg?height=365&width=642&top_left_y=503&top_left_x=707)

Hint. Compare the lengths of segments $AE$ and $EB$.
The midpoint of segment $B E$, through which the line $D F$ passes according to the problem, is denoted as $G$. Segment $F G$ is the midline of triangle $B C E$, which is parallel to side $B C$. In particular, quadrilateral $G B C D$ is also a parallelogram, and therefore segments $E G, G B, D C$, and $A E$ are all congruent.
![](https://cdn.mathpix.com/cropped/2024_04_17_b1dd73eb0c32fed574d9g-4.jpg?height=360&width=1400&top_left_y=1265&top_left_x=334)

The trapezoid $A B C D$ can thus be divided into four triangles $A E D, D C E, E G C$, and $G B C$ with the same area (the first three triangles are even congruent). The area of the trapezoid is therefore equal to four times the area of triangle $C D E$, i.e.,

$$
4 \cdot 3 = 12 \text{ cm}^2.
$$

Note. Triangles $D F C$ and $G F E$ are congruent, so parallelogram $A E C D$ has the same area as triangle $A G D$, and that is congruent to triangle $E B C$. (In both cases, the congruence of the triangles can be justified in several ways, for example, by the SAS theorem.) The area of trapezoid $A B C D$ is therefore equal to twice the area of parallelogram $A E C D$, and that is equal to twice the area of triangle $C D E$.

It also follows from the above that the area of triangle $E B C$ is four times the area of triangle $D F C$, and that is equal to half the area of triangle $C D E$.