vyfukas wrote: Given a triangle $ABC$ let $D$, $E$, $F$ be orthogonal projections from $A$, $B$, $C$ to the opposite sides respectively. Let $X$, $Y$, $Z$ denote midpoints of $AD$, $BE$, $CF$ respectively. Prove that perpendiculars from $D$ to $YZ$, from $E$ to $XZ$ and from $F$ to $XY$ are concurrent. By Karno theorem in triangle $XYZ$ it is enough to prove: $FX^2-EX^2+EZ^2-DZ^2+DY^-FY^2=0$ $(1)$. Using: $FX^2-EX^2=\frac{1}{4}(AF^2+FD^2-AE^2-ED^2).$ $(1)$ is easy.

IMI-Mathboy wrote: By Karno theorem in triangle $XYZ$ it is enough to prove: $FX^2-EX^2+EZ^2-DZ^2+DY^-FY^2=0$ $(1)$. Using: $FX^2-EX^2=\frac{1}{4}(AF^2+FD^2-AE^2-ED^2).$ $(1)$ is easy. What is Karno theorem?

(KARNO).Let $A_1,B_1$ and $C_1$ be points on the sides $BC,AC$ and $AB$ of $\triangle{ABC}$. The lines through $A_1,B_1$ and $C_1$, and perpendicular to $BC,AC$ and $AB$ respectively are concurrent if only if $AB_1^2-B_1C^2+CA_1^2-A_1B^2+BC_1^2-AC_1^2=0$

To avoid calculations we can use the following lemma: Let $ABC, XYZ$ be two triangles. We say that the perpendiculars from $A$ to $YZ$, from $B$ to $XZ$ and from $C$ to $XY$ are concurrent iff the perpendiculars from $X$ to $BC$, from $Y$ to $AC$ and from $Z$ to $AB$ are concurrent. İn some instances, like this one is far easier to prove the reverse. The proof of lemma uses the before mentıoned Carnot lemma. Now to problem: we want to show that perpendiculars: from $X$ to $EF$, from $Y$ to $FD$ and from $Z$ to $DE$ are concurrent. Let $H$ be the orthocenter of $\Delta DEF$ and $O$ the circumcenter of $\Delta ABC$. Because the radii from $A, B, C$ of circle $\odot (ABC)$ are perpendicular to $EF, FD,DE$ respectively, we infer that our perpendiculars pass through the midpoint of $OH$. Best regards, sunken rock

Using the Carnot's theorem but nice trigono calculation.Let's change the problem into following perspective, Let $I_a$, $I_b$ and $I_c$ be the excenters of the $\triangle ABC$. And let $D, E,F$ be the midpoint of $AI_a$, $BI_b$, $CI_c$, respectively. Then prove that perpendiculars from $D,E,F$ to the sides $BC, AC, AB$ are concurrent. Proof: Let's $2 \alpha, 2\beta, 2\gamma$ be the angles of $\triangle ABC$, and let $R$ be the circumradius. Then we have that: $I_bB=4Rcos(\gamma)cos(\beta)$ $\implies$ $BE=2Rcos(\gamma)cos(\beta)$. By the cosine's theorem on the $ABE$ we have that: $AE^2=AB^2+BE^2-2AB*BE*cos(\alpha))$. By the cosine's theorem on the $ACE$ we have that: $CE^2=CB^2+BE^2-2CB*BE*cos(\alpha))$. Thus: $AE^2-CE^2=(AB^2-AC^2)+(2BEcos(\alpha)(BC-AC)=(AB^2-AC^2)+4Rcos(\gamma)cos(\beta)cos(\alpha)(BC-AC)$. $\sum{AE^2-CE^2}=\sum{(AB^2-AC^2)}+4Rcos(\gamma)cos(\beta)cos(\alpha)\sum{(BC-AC)}=0$, which means we are done by Carnot's theorem.