If $\triangle DEF$ is inscribed in $\triangle ABC$ such that three subtriangles have equal inradii, find inradius of $\triangle DEF$

I have a triangle $$ABC$$ with inradius $$r$$. Points $$D$$, $$E$$, $$F$$ are chosen on side $$BC$$, $$CA$$, $$AB$$, respectively, such that $$\triangle AFE$$, $$\triangle BDF$$, and $$\triangle CED$$ have same inradius $$r_1$$. Compute the inradius of $$\triangle DEF$$ in terms of $$r$$ and $$r_1$$.

I have an approach in mind by taking the various side lengths as x,y,z and so on but I know for sure it will turn out to be lengthy. Any better method please?

$$\begin{eqnarray*}r_1\left(p_{AFE}+p_{BDF}+p_{CED}\right)&=&2\left([AFE]+[BDF]+[CED]\right)\\&=&2\left([ABC]-[DEF]\right)\\&=&r\,p_{ABC}-r_{DEF}\,p_{DEF}\\&=&r_1\left(p_{ABC}+p_{DEF}\right)\end{eqnarray*}$$ leads to $$r_{DEF}\, p_{DEF} = (r-r_1) p_{ABC} - r_1 p_{DEF}.$$ Now the interesting fact is that the perimeter of $$DEF$$ is fixed by $$r,r_1,p_{ABC}$$.
By considering the distances of $$D,E,F$$ from the tangency points marked with a red cross we have that $$p_DEF$$ equals the perimeter of $$I_A I_B I_C$$, with $$I_A,I_B,I_C$$ being the incenters of $$AEF,BDF,CDE$$. It follows that
$$p_{DEF} = \frac{r-r_1}{r} p_{ABC}$$ and
$$r_{DEF} = r- r_1.$$
Another interesting fact is that $$D,E,F$$ and $$I_A,I_B,I_C$$ belong to the same ellipse.
• How did you compute the perimeter of triangle $I_AI_BI_C$? – saisanjeev Oct 3 '18 at 14:22
• @saisanjeev: $I_A I_B I_C$ is a scaled version of $ABC$. The dilation factor is clearly $\frac{r-r_1}{r}$. – Jack D'Aurizio Oct 3 '18 at 15:42