In the trapezoid $ABCD$, the diagonals intersect at point $M$. The areas of triangles $ABM$ and $CDM$ are 18 and 50 square units, respectively. What is the area of the trapezoid?
Solution. The triangles $A B M$ and $C D M$ are similar because $\angle M A B = \angle M C D$ and $\angle M B A = \angle M D C$, as they are alternate interior angles. Let $m_{1}$ be the altitude from $M$ in triangle $A B M$, and $m_{2}$ be the altitude from $M$ in triangle $C D M$. The similarity ratio of the two triangles is:

$$
k=\frac{A B}{C D}=\frac{m_{1}}{m_{2}}
$$

Then the ratio of their areas is:

$$
k^{2}=\frac{T_{A B M}}{T_{C D M}}=\frac{\frac{A B \cdot m_{1}}{2}}{\frac{C D \cdot m_{2}}{2}}=\frac{18}{50}
$$

From this, $k=\frac{3}{5}$. Since the height of the trapezoid is $m_{1}+m_{2}$, its area is:

$$
\begin{aligned}
T & =\frac{A B+C D}{2}\left(m_{1}+m_{2}\right)=\frac{k \cdot C D+C D}{2}\left(k \cdot m_{2}+m_{2}\right)= \\
& =(k+1)^{2}\left(C D \cdot \frac{m_{2}}{2}\right)=\left(\frac{8}{5}\right)^{2} \cdot 50=128 .
\end{aligned}
$$

Therefore, the area of the trapezoid is 128 units.

![](https://cdn.mathpix.com/cropped/2024_05_02_eda4229d03c79e67e665g-1.jpg?height=234&width=442&top_left_y=913&top_left_x=836)