Let $(I,r)$ the inscribed circle of $\triangle ABC$ The tangent parallel to the side $BC$ intersects the sides $AB,AC$ at the points $B_1,C_1$, forming in this way a triangle $T_1=\triangle AB_1C_1$. Since $B_1C_1 \parallel BC$, we have $\triangle AB_1C_1 \sim \triangle ABC$, with ratio of similarity $m_1=\frac{B_1C_1}{BC}$ Similarly, the other formed triangles $T_2=\triangle A_2BC_2, \ \ T_3 = \triangle A_3B_3C$ are also similar to $\triangle ABC$, with ratio $m_2$ and $m_3$ respectively. The distance of the parallel lines $BC,B_1C_1$ is $2r$ Let $AD$ the altitude of $\triangle ABC$, and $AD_1$ the altitude of $\triangle AB_1C_1$ We bring a parallel line $L$ from $A$ to $BC$. From the parallel lines $L, B_1C_1,BC$ we get: $\frac{AD_1}{AD} = \frac{AB_1}{AB} = m_1 \Rightarrow$ $AD_1 = m_1 AD$ But $DD_1$ is the diameter of $(I,r)$, so $DD_1=2r\Rightarrow AD=AD_1+2r$ $AD = m_1 AD + 2r \Rightarrow$ $(1-m_1)AD = 2r \Rightarrow$ $\frac{AD}{r} = \frac{2}{1-m_1} \ \ \ (1)$ Let $s = \frac{a+b+c}{2}$, the semi-perimeter of $\triangle ABC$ The area of $ABC$ can be expressed in two different ways: $(ABC) = \frac{AD \cdot a}{2} = s \cdot r \Rightarrow$ $\frac{AD}{r} = \frac{2s}{a} \ \ \ (2)$ $(1),(2) \Rightarrow$ $\frac{2}{1-m_1} = \frac{2s}{a}$ So we find that $m_1 = \frac{s-a}{s}$, and similarly $m_2 = \frac{s-b}{s}, m_3 = \frac{s-c}{s}$ The area of $(I,r)$ is $\pi \cdot r^2$. Each of the triangles $T_1,T_2,T_3$ is similar to the triangle $ABC$, so the incircle of $T_k$ is similar to $(I,r)$ with the same ratio of similarity $m_k$. So we have: Inradius of $T_1 = m_1 \cdot r$ Inradius of $T_2 = m_2 \cdot r$ Inradius of $T_3 = m_3 \cdot r$ Hence, the areas of their inscribed circles are $\pi {m_1}^2 r^2,\ \ \pi {m_2}^2 r^2,\ \ \pi {m_3}^2 r^2$ So we want the sum $S = \pi \cdot r^2 \cdot (1+{m_1}^2+{m_2}^2+{m_3}^2)$ The problem is already solved, because $m_1,m_2,m_3$ are functions of $a,b,c$. But I'll try to find another expression, not containing $m_1,m_2,m_3$ ${m_1}^2 = (\frac{s-a}{s})^2 = (1-\frac{a}{s})^2 = 1-2\frac{a}{s}+\frac{a^2}{s^2}$ ${m_2}^2 = (\frac{s-b}{s})^2 = (1-\frac{b}{s})^2 = 1-2\frac{b}{s}+\frac{b^2}{s^2}$ ${m_3}^2 = (\frac{s-c}{s})^2 = (1-\frac{c}{s})^2 = 1-2\frac{c}{s}+\frac{c^2}{s^2}$ $\Rightarrow$ ${m_1}^2 +{m_2}^2 +{m_3}^2 = 3 -2\cdot \frac{a+b+c}{s} +\frac{a^2+b^2+c^2}{s^2}$ ${m_1}^2 +{m_2}^2 +{m_3}^2 = 3 -2\cdot 2 +\frac{a^2+b^2+c^2}{s^2} = -1 +\frac{a^2+b^2+c^2}{s^2}$ Hence, $1+{m_1}^2 +{m_2}^2 +{m_3}^2 = \frac{a^2+b^2+c^2}{s^2}$ Finally, $S = \pi \cdot r^2 \cdot \frac{a^2+b^2+c^2}{s^2}$

Attachments:

Addition $$S =\pi\cdot r^{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{2}}$$ We know that the area of $ABC$ can be expressed as: $$(ABC) = s\cdot r\Rightarrow r=\frac{(ABC)}{s}$$ $$S =\pi\cdot (\frac{(ABC)}{s})^{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{2}}$$ $$S =\pi\cdot (ABC)^{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{4}}$$ We know also Heron's formula: $(ABC) = \sqrt{s(s-a)(s-b)(s-c)}$ $$S =\pi\cdot s(s-a)(s-b)(s-c)\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{4}}$$ $$S =\pi\cdot (s-a)(s-b)(s-c)\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{3}}$$ Because $s=\frac{a+b+c}{2}$, $s-a=\frac{-a+b+c}{2}$, $s-b=\frac{a-b+c}{2}$ and $s-c=\frac{a+b-c}{2}$: $$S =\pi\cdot \frac{(a+b-c)(b+c-a)(c+a-b)(a^{2}+b^{2}+c^{2})}{(a+b+c)^{3}}$$

Obviously the three smaller triangles are similar to triangle $ABC$. We employ barycentric coordinates with reference triangle $ABC$. Since $I=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)$, the ratios of the side lengths to the side lengths of $ABC$ are $$\frac{a+b-c}{a+b+c},\frac{a-b+c}{a+b+c},\frac{-a+b+c}{a+b+c},\frac{a+b+c}{a+b+c}.$$ Summing the squares, the ratio of the final area to the area of the inscribed circle of triangle $ABC$ is just $$\frac{4(a^2+b^2+c^2)}{(a+b+c)^2}.$$ So our answer is just $$\frac{\pi(a^2+b^2+c^2)(a+b-c)(a-b+c)(-a+b+c)}{(a+b+c)^3}.$$