The area and perimeter of a figure $ \mathcal{F} $ will be denoted by $ [\mathcal{F}] $ and $ p(\mathcal{F}) $ respectively. Note that $ r=\frac{2[\triangle AEF]}{p(\triangle AEF)}=\frac{2[\triangle BDF]}{p(\triangle BDF)}=\frac{2[\triangle CDE]}{p(\triangle CDE)}=\frac{2[\triangle ABC]-2[\triangle DEF]}{p(\triangle ABC)+p(\triangle DEF)}= $ $ \frac{R(p(\triangle ABC))-r_0(p(\triangle DEF))}{p(\triangle ABC)+p(\triangle DEF)}=\frac{R \left(\frac{p(\triangle ABC)}{p(\triangle DEF)} \right)-r_0}{\frac{p(\triangle ABC)}{p(\triangle DEF)}+1} $. If incircle of $ \triangle AEF $ touches $ EF,AE,AF $ at $ P,Q,S $, then we have $ FP=FS,EP=EQ $ and thus $ EF=FS+EQ=AF+AE-AS-AQ=AF+AE-2AS $. But since $ A $ is the homothetic center of incircles of $ \triangle AEF $ and $ \triangle ABC $, we get $ AS=\frac{r}{R}(s-a) $. Thus, we have $ EF=AE+AF-\frac{2r}{R}(s-a) $. Similarly, we get $ DF=BF+BD-\frac{2r}{R}(s-b) $ and $ DE=CD+CE-\frac{2r}{R}(s-c) $. Adding, we have $ p(\triangle DEF)=p(\triangle ABC)-\frac{2r}{R}(s-a+s-b+s-c) $ $ =p(\triangle ABC) \left(1-\frac{r}{R} \right) $ $ \implies \frac{p(\triangle ABC)}{p(\triangle DEF)}=\frac{R}{R-r} $. Thus, we get $ r=\frac{\frac{R^2}{R-r}-r_0}{\frac{R}{R-r}+1} $ $ =\frac{R^2-r_0R+r_0r}{2R-r} \implies 2Rr-r^2=R^2-r_0R+r_0r $ $ \implies (R-r)^2=r_0(R-r) $ $ \implies R-r=r_0 $ $ \implies r+r_0=R $. $ \blacksquare $

Suppose the in-circle of triangle AEF touches sides AE, AF at A'. A" respectively. In the same way B',B" and C',C" are defined and let AA'=AA"=u, BB'=BB"=v, CC'=CC"=w. [ABC] denotes the area of triangle ABC. [ABC]=[AEF]+[BDF]+[CDE]+[DEF], that is; R(a+b+c)=r{2u+2v+2w+2(b-u-w)+2(c-u-v)+2(a-v-w)}+r'(a-v-w+b-u-w+c-u-v)...<1> Let I(A),I(B),I(C) be the incenters of triangles AEF ,BDF,CDE respectively. Then by rearranging the quadrilaterals AA'I(A)A", BB'I(B)B", CC'I(C)C", we get a new triangle ABC with inradius r, similar to triangle ABC. Hence 2(u+v+w) / (a+b+c)=r/R...<2>. From equations <1>, <2>, we get (R-r)(R-r-r')=0 Hence R=r+r'. Here r' denotes the inradius of triangle DEF.