Problem 1606

1. Define the problem and notation:
Let $\triangle ABC$ be an acute-angled triangle with points $D, E, F$ on sides $BC, CA, AB$ respectively. The inradii of triangles $\triangle AEF, \triangle BDF, \triangle CDE$ are all equal to $r_0$. Let $r$ and $R$ be the inradii of triangles $\triangle DEF$ and $\triangle ABC$ respectively. We need to prove that $r + r_0 = R$.

2. Introduce the centers of the incircles:
Let $O_1, O_2, O_3$ be the centers of the incircles of $\triangle AEF, \triangle BDF, \triangle CDE$ respectively. The incircles touch $EF, FD, DE$ at points $M, N, L$ respectively. The incircles of $\triangle BDF$ and $\triangle CDE$ touch $BC$ at points $P, Q$ respectively. The incircles of $\triangle CDE$ and $\triangle AEF$ touch $CA$ at points $R, S$ respectively. The incircles of $\triangle AEF$ and $\triangle BDF$ touch $AB$ at points $T, U$ respectively.

3. Establish similarity of triangles:
Since $O_1O_2UT, O_2O_3QP, O_3O_1SR$ are rectangles and $O_1, O_2, O_3$ lie on the internal bisectors of $\angle A, \angle B, \angle C$, it follows that $\triangle ABC$ and $\triangle O_1O_2O_3$ are centrally similar through their common incenter $I$.

4. Determine the similarity coefficient:
The similarity coefficient is the ratio of their perimeters or inradii:
$$\frac{r - r_0}{r} = \frac{O_1O_2 + O_2O_3 + O_3O_1}{a + b + c} = \frac{PQ + RS + TU}{a + b + c}$$
Since $PQ, RS, TU$ are segments on the sides of $\triangle DEF$, we can write:
$$\frac{r - r_0}{r} = \frac{DN + DL + EL + EM + FM + FN}{a + b + c} = \frac{DE + EF + FD}{a + b + c}$$

5. Relate the inradii and perimeters:
On the other hand, we have:
$$r(a + b + c) = r_0(a + b + c + DE + EF + FD) + \varrho(DE + EF + FD)$$
Simplifying, we get:
$$r - r_0 = \frac{(DE + EF + FD)(r_0 + \varrho)}{a + b + c}$$

6. **Substitute and solve for $r$:**
Substituting $r - r_0$ from the similarity coefficient equation into the above equation, we get:
$$\frac{r - r_0}{r} = \frac{DE + EF + FD}{a + b + c}$$
Therefore:
$$r - r_0 = \frac{(DE + EF + FD)(r_0 + \varrho)}{a + b + c}$$
Simplifying further, we find:
$$r = r_0 + \varrho$$

7. Conclusion:
Since $\varrho$ is the inradius of $\triangle DEF$, we have $r = r_0 + R$.

$\blacksquare$

The final answer is $\boxed{ r + r_0 = R }$