In the triangle $ABC$, point $D$ is given on the extension of side $CA$ beyond $A$, and point $E$ is given on the extension of side $CB$ beyond $B$, such that $AB = AD = BE$. The angle bisectors from $A$ and $B$ of triangle $ABC$ intersect the opposite sides at points $A_1$ and $B_1$, respectively. What is the area of triangle $ABC$ if the area of triangle $DCE$ is 9 units, and the area of triangle $A_1CB_1$ is 4 units?
Solution. Let the sides of triangle $ABC$ be $a, b$, and $c$ with the angle at $C$ being $\gamma$. Then the area of the triangle $T = ab \cdot \frac{\sin \gamma}{2}$.

According to the Angle Bisector Theorem, $CA_1 = \frac{ab}{b+c}$ and $CB_1 = \frac{ba}{a+c}$, so the area of triangle $A_1CB_1$ is

$$
T_{1} = \frac{(ab)^2}{(a+c)(b+c)} \cdot \frac{\sin \gamma}{2}
$$

Since $CD = b+c$ and $CE = a+c$, the area of triangle $DCE$ is

$$
T_{2} = (a+c)(b+c) \cdot \frac{\sin \gamma}{2}
$$

From these, it follows that $T^2 = T_1 T_2$, or $T = \sqrt{9 \cdot 4} = 6$ square units.

Based on the work of Balázs Poronyi (Pécs, Janus Pannonius Gymnasium, 10th grade)