Perimeter of orthic triangle If $DEF$ is the orthic triangle of $\triangle ABC$, then prove that
$$\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R} $$
where $r$ and $R$ are the inradius and the circumradius of $\triangle ABC$. 
My attempt is very simple , I put the side length of orthic triangle in terms of $\cos$ and the side length of $\triangle ABC$ but I can't get the required answer. 

$$\sin{2A}+\sin{2B}+\sin2C=2\sin{(A+B)}\cos{(A-B)}+2\sin{C}\cos{C}$$$$=2\sin{C}(\cos{(A-B)}-\cos{(A+B)})=4\sin{A}\sin{B}\sin{C}$$ 
$$\text{Now, }\frac{EF}{\sin A}=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}\implies EF=2R\sin A\cos A=R\sin{2A}\text{ .}$$
$$\therefore \text{Perimeter of }\triangle DEF=4R\sin{A}\sin{B}\sin{C}=\frac{2\Delta}{R}=\frac{r\cdot 2s}{R}$$$$\implies\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R}$$
$\blacksquare$
Explanation: $\Delta=\frac{1}{2}ab\sin C=\frac{1}{2}(2R\sin A)(2R \sin B)\sin C=2R^2\sin{A}\sin{B}\sin{C}$
