A circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Tangents to the circle parallel to the sides of the triangle are contructe. Each of these tangents cuts off a triagnle from $\triangle ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$).
1. Let \((I, r)\) be the inscribed circle of \(\triangle ABC\) with radius \(r\).
2. The tangent parallel to the side \(BC\) intersects the sides \(AB\) and \(AC\) at the points \(B_1\) and \(C_1\), forming the triangle \(T_1 = \triangle AB_1C_1\).
3. Since \(B_1C_1 \parallel BC\), we have \(\triangle AB_1C_1 \sim \triangle ABC\), with the ratio of similarity \(m_1 = \frac{B_1C_1}{BC}\).
4. Similarly, the other formed triangles \(T_2 = \triangle A_2BC_2\) and \(T_3 = \triangle A_3B_3C\) are also similar to \(\triangle ABC\), with ratios \(m_2\) and \(m_3\) respectively.
5. The distance between the parallel lines \(BC\) and \(B_1C_1\) is \(2r\).
6. Let \(AD\) be the altitude of \(\triangle ABC\), and \(AD_1\) be the altitude of \(\triangle AB_1C_1\). We bring a parallel line \(L\) from \(A\) to \(BC\).
7. From the parallel lines \(L\), \(B_1C_1\), and \(BC\), we get:
   \[
   \frac{AD_1}{AD} = \frac{AB_1}{AB} = m_1 \Rightarrow AD_1 = m_1 AD
   \]
8. But \(DD_1\) is the diameter of \((I, r)\), so \(DD_1 = 2r \Rightarrow AD = AD_1 + 2r\).
9. Therefore,
   \[
   AD = m_1 AD + 2r \Rightarrow (1 - m_1)AD = 2r \Rightarrow \frac{AD}{r} = \frac{2}{1 - m_1} \quad (1)
   \]
10. Let \(s = \frac{a + b + c}{2}\), the semi-perimeter of \(\triangle ABC\).
11. The area of \(\triangle ABC\) can be expressed in two different ways:
    \[
    (\triangle ABC) = \frac{AD \cdot a}{2} = s \cdot r \Rightarrow \frac{AD}{r} = \frac{2s}{a} \quad (2)
    \]
12. From equations (1) and (2), we get:
    \[
    \frac{2}{1 - m_1} = \frac{2s}{a}
    \]
13. Solving for \(m_1\), we find:
    \[
    m_1 = \frac{s - a}{s}
    \]
    Similarly, we find:
    \[
    m_2 = \frac{s - b}{s}, \quad m_3 = \frac{s - c}{s}
    \]
14. The area of \((I, r)\) is \(\pi r^2\).
15. Each of the triangles \(T_1, T_2, T_3\) is similar to the triangle \(\triangle ABC\), so the incircle of \(T_k\) is similar to \((I, r)\) with the same ratio of similarity \(m_k\). Thus, we have:
    \[
    \text{Inradius of } T_1 = m_1 \cdot r, \quad \text{Inradius of } T_2 = m_2 \cdot r, \quad \text{Inradius of } T_3 = m_3 \cdot r
    \]
16. Hence, the areas of their inscribed circles are:
    \[
    \pi m_1^2 r^2, \quad \pi m_2^2 r^2, \quad \pi m_3^2 r^2
    \]
17. So we want the sum:
    \[
    S = \pi r^2 (1 + m_1^2 + m_2^2 + m_3^2)
    \]
18. We express \(m_1, m_2, m_3\) in terms of \(a, b, c\):
    \[
    m_1^2 = \left(\frac{s - a}{s}\right)^2 = \left(1 - \frac{a}{s}\right)^2 = 1 - 2\frac{a}{s} + \frac{a^2}{s^2}
    \]
    \[
    m_2^2 = \left(\frac{s - b}{s}\right)^2 = \left(1 - \frac{b}{s}\right)^2 = 1 - 2\frac{b}{s} + \frac{b^2}{s^2}
    \]
    \[
    m_3^2 = \left(\frac{s - c}{s}\right)^2 = \left(1 - \frac{c}{s}\right)^2 = 1 - 2\frac{c}{s} + \frac{c^2}{s^2}
    \]
19. Adding these, we get:
    \[
    m_1^2 + m_2^2 + m_3^2 = 3 - 2\left(\frac{a + b + c}{s}\right) + \frac{a^2 + b^2 + c^2}{s^2}
    \]
    \[
    m_1^2 + m_2^2 + m_3^2 = 3 - 2 \cdot 2 + \frac{a^2 + b^2 + c^2}{s^2} = -1 + \frac{a^2 + b^2 + c^2}{s^2}
    \]
20. Hence,
    \[
    1 + m_1^2 + m_2^2 + m_3^2 = \frac{a^2 + b^2 + c^2}{s^2}
    \]
21. Finally,
    \[
    S = \pi r^2 \cdot \frac{a^2 + b^2 + c^2}{s^2}
    \]

The final answer is \( \boxed{ \pi r^2 \cdot \frac{a^2 + b^2 + c^2}{s^2} } \).