Mathematical Reflections 2012, Issue 5 - Problem J242

Problem:
Let $ABC$ be a triangle and let $D,E,F$ be the feet of the altitudes from $A,B,C$ to the
sides $BC,CA,AB$, respectively. Let $X,Y,Z$ be the midpoints of segments $EF,FD,DE$ and let $x,y,z$ be the perpendiculars from $X,Y,Z$ to $BC,CA$, and $AB$, respectively. Prove that the lines $x,y,z$ are concurrent.

Proposed by Cosmin Pohoata.

Solution:

Obviously, the lines $x,y,z$ are parallel to the altitudes $AD,BE,CF$ respectively. Since the orthocenter of the triangle $ABC$ is the incenter of the orthic triangle $DEF$, the lines $AD,BE,CF$ are bisectrices of the angle $D,E,F$ of the triangle $DEF$ respectively. Moreover, since $X,Y,Z$ are midpoints of the segments $EF,FD,DE$, then $XY,YZ,ZX$ are parallel to $DE,EF,FD$. The angle between the line $XY$ and the line $x$ and the angle $ADE$ are equal (alternate interior angles) and so are the angle between the line $XZ$ and the line $x$ and the angle $ADF$ (alternate exterior angles), so the line $x$ is the bisectrix of the angle $X$ of the triangle $XYZ$. Likewise, the lines $y$ and $z$ are bisectrices of the angles $Y,Z$ respectively. So the lines $x,y,z$ are concurrent since their point of intersection is the incenter of the triangle $XYZ$.