I have a really ugly trigonometric solution.. We use the fact that the sides of the orthic triangle are $ a \cos A, b \cos B$ and $ c \cos C$ which is easy to prove. From the given condition we have $3a \cos A = b \cos B + c \cos C$ Hence $3 \sin A \cos A = \sin B \cos B + \sin C \cos C$ (since $a = 2R \sin A$ ) $=> -3 \sin (B+C) \cos (B+C) = \sin B \cos B + \sin C \cos C$ $=> -6 \sin (B+C) \cos (B+C) = \sin (2B) + \sin (2C)$ $=> -6 \sin (B+C) \cos (B+C) = 2 \sin (B+C)\cos (B-C)$ $ => -3 \cos (B+C) = \cos (B-C) $ $ => -3 \cos B \cos C + 3 \sin B \sin C = \cos B \cos C +\sin B \sin C$ $=> \sin B \sin C = 2 \cos B \cos C => -\cos B \cos C = \cos (B+C)$ $=> \cos A = \cos B \cos C$ Note that $ZE = CE \cdot \cos (\angle ZEC) = CE \cos B = a \cos C \cos B = a \cos A = FE$ Also $\angle ZEF_1 = 180 - \angle FED = 2B$ and since $\angle ZEC = B$ we have that $EC$ is the perpendicular bisector of $ZF_1$ in isosceles triangle $ZEF_1$. Similarly $BF$ is the perpendicular bisector of isosceles triangle $E_1FY$. Since $F_1E =EF = ZE$ we have that $\angle FZF' = 90$ Similarly $\angle E_1YE = 90 => \angle BZF_1 = \angle CYE_1 \leftrightarrow \angle BZF = \angle CYE$ Note that since $FZ \perp F_1Z$ and $F_1Z \perp AC$ we have that $FZ$ is parallel to $AC$. Hence $\angle BFZ = \angle BAC$ Similarly $EY$ is parallel to $AB$ and therefore $\angle YEC = \angle BAC = \angle BFZ$. It is enough to prove that $\frac{BF}{CE} = \frac {FZ}{EY}$ which will imply $BFZ \sim CEY$ and we will be done. Note that $BF = a \cos B$ and $CE = a\cos C$. Also since $FZF_1$ is a right triangle and $\angle FF_1Z = \angle EF_1Z = 90 - B$ we have that $FZ = FF_1 \cos B = 2 EF \cos B = 2 a \cos A \cos B$ Similarly on triangle $E_1EY$ we get $EY = 2a \cos A \cos C$. Therefore $\frac {FZ}{EY} = \frac {\cos B}{\cos C} = \frac {BF}{CE}$ which implies the desired result.

Jutaro wrote: Let $ABC$ be an acute triangle and $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. Call $Y$ and $Z$ the feet of the perpendicular lines from $B$ and $C$ to $FD$ and $DE$, respectively. Let $F_1$ be the symmetric of $F$ with respect to $E$ and $E_1$ be the symmetric of $E$ with respect to $F$. If $3EF=FD+DE$, prove that $\angle BZF_1=\angle CYE_1$. We don`t need some heavy computing .

Hint

by a little(not hard or ugly) tc,we can draw the following conclusion: the two conditions$3EF=FD+DE$ and $\angle BZF_1=\angle CYE_1$ are both equivalent to $(b^2+a^2-c^2)(a^2+c^2-b^2)=2a^2(b^2+c^2-a^2)$ where $a,b,c$ are three sides of triangle $ABC$.

mahanmath wrote: Jutaro wrote: Let $ABC$ be an acute triangle and $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. Call $Y$ and $Z$ the feet of the perpendicular lines from $B$ and $C$ to $FD$ and $DE$, respectively. Let $F_1$ be the symmetric of $F$ with respect to $E$ and $E_1$ be the symmetric of $E$ with respect to $F$. If $3EF=FD+DE$, prove that $\angle BZF_1=\angle CYE_1$. We don`t need some heavy computing .

Hint

nice it's how i solved it too now it's easy to get $\angle BZF_{1}=\angle CYE_{1}=270-\angle B-\angle C$ i have a question though when i want refer to what you did for lenghts of tangent lines on the IMO can i just use it without a comment or is that some theorem i have to refer to?