$D$ is a point on the side $A C$ of triangle $A B C$. $A C=18 \mathrm{~cm}$. We know that $A D=5 \mathrm{~cm} . D$ is 4 cm from side $A B$, and 5 cm from side $C B$. Calculate the area of triangle $A B C$.
According to the notation in the figure:

$$
\begin{aligned}
& AC=18 \text{ cm} \\
& AD=5 \text{ cm} \\
& DE=4 \text{ cm} \\
& DF=5 \text{ cm}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a35419db5d41677460d5g-1.jpg?height=247&width=621&top_left_y=500&top_left_x=730)

The majority of solvers calculated the value of angle $\alpha$ at vertex $A$ and angle $\gamma$ at vertex $C$ using a trigonometric function, since $\triangle AED$ and $\triangle DFC$ are right-angled triangles. For example:

$$
\begin{array}{ll}
\sin \alpha=\frac{4}{5}, & \sin \beta=\frac{5}{13} \\
\cos \alpha=\frac{3}{5}, & \cos \beta=\frac{12}{13}
\end{array}
$$

Then they used the formula:

$$
t=\frac{b^{2} \sin \alpha \sin \gamma}{2 \sin (\alpha+\gamma)}=\frac{b^{2} \sin \alpha \sin \gamma}{2(\sin \alpha \cos \gamma+\cos \alpha \sin \gamma)}
$$

With minimal calculation, they obtained the area as $\frac{360}{7} \text{ cm}^2 \approx 51.43 \text{ cm}^2$.

Regardless of the method used to reach the correct result, solvers received 2 points. However, we now present a very elegant and short solution, submitted by Róbert Bagyinszki (Békéscsaba, Sebes Gy. Közg. Szki., II. o. t).

Using the notation in the figure:

$AE=3 \text{ cm}, FC=12 \text{ cm}$. Let the foot of the altitude from $A$ to $AC$ be $T$. Let $AT=x$, $TC=18-x$, and $BT=m$. From the similar triangles $\triangle ADE$ and $\triangle ABT$:

$$
\frac{4}{3}=\frac{m}{x}
$$

From the similar triangles $\triangle CFD$ and $\triangle BTC$:

$$
\frac{5}{12}=\frac{m}{18-x}
$$

Substituting the value of $x$ from (1):

$$
m=\frac{40}{7}, \quad \text{and thus} \quad T_{ABC}=\frac{18 \cdot m}{2}=\frac{18 \cdot \frac{40}{7}}{2}=\frac{360}{7} \text{ cm}^2
$$