Perimeter of orthic triangle

Question

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.

Answer 1

Let $AD$, $BE$ and $CF$ be altitudes of an acute-angled $\Delta ABC$.

Thus, since $BCEF$ is cyclic, we obtain that $\Delta AEF\sim\Delta ABC,$ which gives $$\frac{FE}{BC}=\frac{AF}{AC}$$ or in the standard notation $$\frac{FE}{a}=\cos\alpha.$$ Id est, $$\frac{P_{\Delta DEF}}{P_{\Delta ABC}}=\frac{\sum\limits_{cyc}a\cos\alpha}{a+b+c}=\frac{\sum\limits_{cyc}\frac{a(b^2+c^2-a^2)}{2bc}}{a+b+c}=\frac{\sum\limits_{cyc}a^2(b^2+c^2-a^2)}{2abc(a+b+c)}=$$ $$=\frac{\sum\limits_{cyc}(2a^2b^2-a^4)}{2abc(a+b+c)}=\frac{16S^2}{2abc(a+b+c)}=\frac{\frac{2S}{a+b+c}}{\frac{abc}{4S}}=\frac{r}{R}.$$ I used the cyclic sum: $$\sum_{cyc}a^2(b^2+c^2-a^2)=a^2(b^2+c^2-a^2)+b^2(c^2+a^2-b^2)+c^2(a^2+b^2-c^2)=$$ $$=2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4=\sum_{cyc}(2a^2b^2-a^4).$$ Also, $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=$$ $$=\sqrt{\frac{a+b+c}{2}\cdot\frac{a+b-c}{2}\cdot\frac{a-b+c}{2}\cdot\frac{-a+b+c}{2}}=$$ $$=\sqrt{\frac{1}{16}(2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4)}.$$

Answer 2

$$\sin{2A}+\sin{2B}+\sin2C=2\sin{(A+B)}\cos{(A-B)}+2\sin{C}\cos{C}$$$$=2\sin{C}(\cos{(A-B)}-\cos{(A+B)})=4\sin{A}\sin{B}\sin{C}$$

$$\text{Now, }\frac{EF}{\sin A}=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}\implies EF=2R\sin A\cos A=R\sin{2A}\text{ .}$$

$$\therefore \text{Perimeter of }\triangle DEF=4R\sin{A}\sin{B}\sin{C}=\frac{2\Delta}{R}=\frac{r\cdot 2s}{R}$$$$\implies\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R}$$

$\blacksquare$

Explanation: $\Delta=\frac{1}{2}ab\sin C=\frac{1}{2}(2R\sin A)(2R \sin B)\sin C=2R^2\sin{A}\sin{B}\sin{C}$