## Task B-4.3.

A circle with center at point $S$ is inscribed in triangle $A B C$. The circle touches the sides of the triangle at points $D \in \overline{B C}, E \in \overline{A C}$ and $F \in \overline{A B}$. If $|A E|=3 \mathrm{~cm},|C S|=2 \sqrt{7} \mathrm{~cm}$, and the measure of the angle at vertex $A$ is $60^{\circ}$, calculate the area of triangle $A B C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8b541dbeede02a7e62f7g-17.jpg?height=800&width=600&top_left_y=431&top_left_x=725)

Since $CS$ is the angle bisector, $\angle ECS = \angle DCS$, $\angle CES = \angle CDS = 90^\circ$, and $\overline{CS}$ is a common side, so triangles $CES$ and $CDS$ are congruent by the SAS (Side-Angle-Side) theorem. Therefore, $|CE| = |CD| = y$.

Similarly, we conclude that $|AF| = |AE| = 3$, $|BF| = |BD| = x$.

Let $r = |ES|$ be the radius of the inscribed circle. Since $AS$ is the angle bisector of $\angle EAF$, we have $\angle EAS = 30^\circ$. From triangle $AES$, which is half of an equilateral triangle, it follows that $r = 3 \tan 30^\circ = \sqrt{3} \text{ cm}$.

Applying the Pythagorean theorem to the right triangle $CES$, we get

$y^2 = (2\sqrt{7})^2 - 3^2$, hence $y = 5 \text{ cm}$, and $|AC| = 8 \text{ cm}$.

The semiperimeter of triangle $ABC$ is $s = \frac{8 + 5 + x + 3 + x}{2} = 8 + x$. The length $x$ is calculated using the formulas for the area $P = r \cdot s$ and $P = \frac{1}{2} ab \sin \alpha$:

$$
\begin{gathered}
r s = \frac{1}{2} |AB| \cdot |AC| \sin 60^\circ \\
\sqrt{3}(8 + x) = \frac{1}{2} \cdot 8 \cdot (3 + x) \cdot \frac{\sqrt{3}}{2} \\
8 + x = 6 + 2x \\
x = 2 \text{ cm}
\end{gathered}
$$

The sides of the triangle are $8 \text{ cm}$, $7 \text{ cm}$, and $5 \text{ cm}$, and the area is $P = r s = \sqrt{3}(8 + 2) = 10\sqrt{3} \text{ cm}^2$.