In trapezoid $ABCD$, the bases are $AB$ and $CD$, and the intersection point of the diagonals is $M$. The diagonal $AC$ bisects the angle $BAD$, $AM = BC$, and $BM = CD$. Determine the angles of the trapezoid.
The angles $\angle CAB$ and $\angle ACD$ lying on the bases of the trapezoid are alternate interior angles, hence they are equal. Therefore, the triangle $ACD$ is isosceles. A known property of the trapezoid is that the areas of triangles $AMD$ and $BCM$ are equal. According to the problem's conditions and the previously mentioned angle equality, $AM = BC$, and $BM = CD = DA$. Thus, not only do the areas of triangles $AMD$ and $BCM$ match, but two pairs of their sides also match (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_619c8fe3515120282dc0g-1.jpg?height=299&width=412&top_left_y=352&top_left_x=843)

Figure 1

We will prove that these two triangles are congruent.

To do this, we first examine how many non-congruent triangles there are with two pairs of equal sides and equal area. If the area is given, then we also know the heights corresponding to the given sides. If we take one of the given sides $(PQ)$ and draw a line parallel to this side at the known height, then it is certain that the third vertex of the triangle can only be located on this parallel line. The third vertex $(R)$ is also on the circumference of a circle with the radius equal to the length of the other given side, centered at one of the endpoints of the given segment. The parallel line and the circumference intersect at most at two points ($R_1, R_2$) as shown in Figure 2. (Drawing the other parallel line, swapping the endpoints, and the traversal does not lead to a different solution.)

![](https://cdn.mathpix.com/cropped/2024_05_02_619c8fe3515120282dc0g-1.jpg?height=434&width=529&top_left_y=1168&top_left_x=782)

Figure 2

The segments $R_1Q$ and $R_2Q$ are symmetric with respect to the perpendicular line to $PQ$ through the center of the circle, so $\angle PQR_1 = \angle PQR_2$, meaning that if the angles $\angle PQR_1$ and $\angle PQR_2$ are not equal, they sum up to $180^\circ$.

Now let's return to the examination of triangles $AMD$ and $BCM$. Can the angles $\alpha$ and $\gamma$ marked on the first figure be supplementary? Based on the placement of the angles, $\alpha = \angle ACD < \angle BCD$ and $\gamma = \angle DBC < \angle ABC$, so in this case,

$$
180^\circ = \alpha + \gamma < \angle BCD + \angle ABC
$$

which is a contradiction because the angles at $B$ and $C$ of the trapezoid $ABCD$ sum up to $180^\circ$. Therefore, the angles $\alpha$ and $\gamma$ are equal.

From here, the solution can be completed in a few steps. We have thus shown that $\triangle DAM \cong \triangle MBC$. The angles $\angle AMD$ and $\angle BMC$ are vertical angles, so they are equal, i.e., $\angle ADM = \angle BMC = \angle AMD = \angle BCM$. Therefore, these two triangles are also isosceles. The triangle $DMC$ is also isosceles, and its exterior angle at $D$ is $\angle DMA$, which is $2\alpha$ in size. For the angles of the isosceles triangle $AMD$,

$$
\angle MAD + \angle AMD + \angle MDA = \alpha + 2\alpha + 2\alpha = 180^\circ
$$

From this, $\alpha = 36^\circ$, and finally, the angles of the trapezoid $ABCD$ are $72^\circ, 72^\circ, 108^\circ, 108^\circ$.

Based on the work of Zsigmond Fleiner (Budapest Fazekas M. Gyak. Ált. Isk. és Gimn., 9th grade)