Given the sides of $\Delta ABC$ are: $a, b, c$. Inside the triangle, we draw lines parallel to the sides at a distance $d$. The task is to calculate the area of the resulting triangle!

Source: Jenő Szép (Zrínyi Miklós g. VIII o. t. Bp. VIII.)
Let the vertices of the inner triangle be $A^{\prime}, B^{\prime}, C^{\prime}$. Since $A^{\prime}$ is equidistant from the sides $AB$ and $AC$, it lies on the angle bisector of $\alpha$; and above this bisector, it forms the angle $\alpha^{\prime}$. Similarly, $B^{\prime}$ lies on the angle bisector of $\beta$, and $C^{\prime}$ lies on the angle bisector of $\gamma$, and these bisectors also form the angles $\beta^{\prime}$ and $\gamma^{\prime}$. From this, it follows that the incenter $I$ is common to both triangles, and the two similar triangles are similarly situated with respect to $I$, so the ratio of their areas is equal to the ratio of the squares of the corresponding sides, and also to the ratio of the squares of the radii of the inscribed circles.

![](https://cdn.mathpix.com/cropped/2024_05_02_556ea430b61d71902f30g-1.jpg?height=412&width=499&top_left_y=388&top_left_x=802)

If the area of $\triangle ABC$ is $t$, and the area of $\triangle A^{\prime}B^{\prime}C^{\prime}$ is $t^{\prime}$, and the radius of the inscribed circle for the former is $\varrho$, and for the latter is $\varrho^{\prime}=\varrho-d$, then

$$
t^{\prime}: t=(\varrho-d)^{2}: \varrho^{2} \quad \text { or } \quad t^{\prime}: t=\left(1-\frac{d}{\varrho}\right)^{2}: 1
$$

Since

$$
\begin{aligned}
& \varrho=\frac{2 t}{a+b+c}=\frac{2 t}{2 s}=\frac{t}{s} \\
& t^{\prime}=t\left(1-\frac{d s}{t}\right)^{2}=\frac{(t-d s)^{2}}{t}
\end{aligned}
$$

where

$$
t=\sqrt{s(s-a)(s-b)(s-c)}
$$