The "excribed triangle" to a side of a triangle has vertices at the points where the circle tangent to the given side and the extensions of the other two sides touches these lines. The sides of the triangle are $a, b, c$. What is the ratio of the areas of the "excribed triangles" to the sides $a$ and $b$?
Solution. Let the vertices and angles of the triangle be denoted in the usual way by $A, B, C$, and $\alpha, \beta, \gamma$, respectively. The vertices of the triangle inscribed to the side $a$ are denoted by $P, Q, R$ as shown in the figure, the center of the excircle by $O_{A}$, and its radius by $r_{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4af1cdacb8576652d634g-1.jpg?height=590&width=553&top_left_y=304&top_left_x=775)

The quadrilaterals $O_{A} R B P$ and $O_{A} Q C R$ are cyclic, since the angles at $P$ and $R$, and at $Q$ and $R$, respectively, are right angles, because the tangents to the excircle are perpendicular to the radii at the points of tangency. Therefore, $P O_{A} R \varangle = A B C \varangle = \beta$ and $Q O_{A} R \varangle = A C B \varangle = \gamma$. Since $\beta + \gamma < 180^{\circ}$, $O_{A}$ and $R$ are on different sides of the line $P Q$. Thus, the area of the triangle inscribed to the side $a$ is

$$
T_{P Q R} = T_{P O_{A} R} + T_{R O_{A} Q} - T_{P O_{A} Q}
$$

Using the fact that $O_{A} P = O_{A} Q = O_{A} R = r_{a}$, and that $\beta + \gamma = 180^{\circ} - \alpha$, we get

$$
T_{P Q R} = \frac{r_{a}^{2}}{2} (\sin \beta + \sin \gamma - \sin (\beta + \gamma)) = \frac{r_{a}^{2}}{2} (\sin \beta + \sin \gamma - \sin \alpha)
$$

Similarly, the area of the triangle inscribed to the side $b$ (where $r_{b}$ is the radius of the excircle to the side $b$) can be written as: $\frac{r_{b}^{2}}{2} (\sin \alpha + \sin \gamma - \sin \beta)$.

Introducing the notation $2 s = a + b + c$, and using the sine theorem and the relationship for the radii of the excircles $r_{a}(s - a) = r_{b}(s - b) \left(= T_{A B C}\right)$ (a proof of this can be found, for example, in Kiss Gy.: What is Good to Know About Triangles, KöMaL 2002/3, pp. 130-139), we get that the ratio of the areas of the two inscribed triangles is:

$$
\frac{\frac{r_{a}^{2}}{2} (\sin \beta + \sin \gamma - \sin \alpha)}{\frac{r_{b}^{2}}{2} (\sin \alpha + \sin \gamma - \sin \beta)} = \frac{\frac{1}{(s-a)^{2}}}{\frac{1}{(s-b)^{2}}} \cdot \frac{s-a}{s-b} = \frac{s-b}{s-a}
$$