## Task B-2.4.

A triangle $A B C$ is given. Point $D$ is on side $\overline{A B}$, and point $E$ is on side $\overline{B C}$ of triangle $A B C$ such that $|A D|=3 \mathrm{~cm},|B D|=7 \mathrm{~cm},|B E|=8 \mathrm{~cm},|D E|=5 \mathrm{~cm}$, and $\varangle B A C=\varangle D E B$. What is the area of quadrilateral $A D E C$?
## First solution.

Let's sketch the triangle and mark all the given elements.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-12.jpg?height=280&width=423&top_left_y=628&top_left_x=248)

Triangles $A B C$ and $E B D$ are similar by the $K K$ similarity criterion because they have two pairs of congruent angles ( $\angle B A C = \angle D E B$, and $\angle C B A$ is common to both).

The similarity coefficient of these triangles is $k = \frac{|A B|}{|B E|} = \frac{10}{8} = \frac{5}{4}$.

We calculate the area of triangle $E B D$ using Heron's formula:

$P(\triangle E B D) = \sqrt{10 \cdot (10 - 8) \cdot (10 - 7) \cdot (10 - 5)} = \sqrt{10 \cdot 2 \cdot 3 \cdot 5} = 10 \sqrt{3} \, \text{cm}^2$.

For the areas of similar triangles, it holds that $\frac{P(\triangle A B C)}{P(\triangle E B D)} = k^2$.

The area of triangle $A B C$ is:

$P(\triangle A B C) = P(\triangle E B D) \cdot k^2 = 10 \sqrt{3} \cdot k^2 = 10 \sqrt{3} \cdot \frac{25}{16} = \frac{125 \sqrt{3}}{8} \, \text{cm}^2$.

The area of quadrilateral $A D E C$ is:

$P(A D E C) = P(\triangle A B C) - P(\triangle E B D) = \frac{125 \sqrt{3}}{8} - 10 \sqrt{3} = \frac{45 \sqrt{3}}{8} \, \text{cm}^2$.