In triangle $ABC$, points $D$ and $E$ lie on side $AB$, with $D$ being closer to $A$. $AD=100 \mathrm{~m}, EB=200 \mathrm{~m}$, $\angle ACD=30^{\circ}, \angle DCE=50^{\circ}$, and $\angle ECB=35^{\circ}$. What is the length of segment $DE$?
Let's introduce the following notations: $BC=a, CA=b, ED=x, EC=d, CD=e$, and let the height from point $C$ be $m$.

![](https://cdn.mathpix.com/cropped/2024_05_02_585d0a62d800e11101d8g-1.jpg?height=689&width=1347&top_left_y=260&top_left_x=389)

Write the double area of the three sub-triangles and the entire $\triangle ABC$ in two ways: using two sides and the sine of the included angle, and using the base and the corresponding height, then set the expressions equal to each other:

$$
\begin{aligned}
200 m & =a d \sin 35^{\circ} \\
x m & =d e \sin 50^{\circ} \\
100 m & =b e \sin 30^{\circ} \\
(300+x) m & =a b \sin 115^{\circ}
\end{aligned}
$$

Form the product of equations (2) and (4), and the product of equations (1) and (3) - i.e., the products of the sides opposite each other - then divide the former product by the latter (which we can clearly do, since no area is 0). Thus, the auxiliary variables $a, b, d, e$ drop out, and we obtain the following equation:

$$
\begin{gathered}
\frac{x(300+x)}{20000}=\frac{\sin 50^{\circ} \cdot \sin 115^{\circ}}{\sin 35^{\circ} \cdot \sin 30^{\circ}} \\
x^{2}+300 x-40000 \frac{\sin 50^{\circ} \cdot \sin 65^{\circ}}{\sin 35^{\circ}}=0
\end{gathered}
$$

From the signs of the coefficients, we see that the equation has one positive and one negative root, and only the positive root fits our problem and thought process:

$$
x=-150+\sqrt{22500+40000 \frac{\sin 50^{\circ} \sin 65^{\circ}}{\sin 35^{\circ}}}=116.3 \text{ m}.
$$