The given points of tangency of the incircle of $\triangle ABC$ on the sides are $A_{1}, B_{1}, C_{1}$. The sides, angles, and area of $\triangle A_{1} B_{1} C_{1}$ need to be calculated!
$1^{0}$. Let $I$ be the center of the inscribed circle of $\triangle ABC$. The angle $\angle B_{1}IC_{1}=180^{\circ}-\alpha$. Let the radius of the inscribed circle of $\triangle ABC$ be $r$. The triangle $\triangle A_{1}B_{1}C_{1}$ is inscribed in a circle of radius $r$; therefore,

$$
B_{1}C_{1}=2r \sin \frac{1}{2}(180^{\circ}-\alpha)=2r \cos \frac{\alpha}{2}=\frac{2t}{s} \cos \frac{\alpha}{2}
$$

Here, $t$ represents the area of $\triangle ABC$, and $s$ represents half the perimeter of $\triangle ABC$. Similarly,

$$
C_{1}A_{1}=\frac{2t}{s} \cos \frac{\beta}{2}, \quad A_{1}B_{1}=\frac{2t}{s} \cos \frac{\gamma}{2}
$$

$2^{0}$. The angles of $\triangle A_{1}B_{1}C_{1}$ are inscribed angles in a circle of radius $r$. The inscribed angle $\alpha_{1}=\angle B_{1}A_{1}C_{1}$ corresponds to the central angle $\angle B_{1}IC_{1}=180^{\circ}-\alpha$. Therefore,

$$
\alpha_{1}=90^{\circ}-\frac{\alpha}{2}, \quad \beta_{1}=90^{\circ}-\frac{\beta}{2}, \quad \gamma_{1}=90^{\circ}-\frac{\gamma}{2}
$$

$3^{0}$. Let $t_{1}$ be the area of $\triangle A_{1}B_{1}C_{1}$. The radius of its circumscribed circle is $r$, so

$$
\begin{gathered}
t_{1}=\frac{a_{1}b_{1}c_{1}}{4r}=\frac{(2r)^{3}}{4r} \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}=2r^{2} \sqrt{\frac{s(s-a)}{bc}} \sqrt{\frac{s(s-b)}{ca}} \sqrt{\frac{s(s-c)}{ab}}= \\
=\frac{2r^{2}}{abc} s \sqrt{s(s-a)(s-b)(s-c)}=\frac{2r^{2}}{4Rt} \cdot s \cdot t=\frac{2r \cdot(rs)t}{4Rt} \\
t_{1}=\frac{r}{2R} t
\end{gathered}
$$

Here, $t$ represents the area of $\triangle ABC$, and $R$ represents the radius of its circumscribed circle.