# A geometry problem about the nine-point-circle

In the acute angled triangle $ABC$, the midpoints of the sides $BC$, $CA$ and $AB$ are $D$, $E$ and $F$, respectively. The foot of the altitude of the triangle starting from $C$ is $T_1$. On some line $l$, passing through point $C$ but not containing $T_1$, the feet of the perpendiculars starting from $A$ and $B$ are $T_2$ and $T_3$, respectively. Prove that the circle $DEF$ passes through the center of the circle $T_1 T_2 T_3$.

I found that $ABC \sim T_1 T_2 T_3$ , and now we are learning the Feuerbach's circle.

(illustrated in geogebra)

• Try to find many cyclic quadrilaterals and exploit the fact that the circumcircle of $DEF$ is the nine-point circle of $ABC$, hence it goes through the feet of the altitudes of $ABC$, too. Oct 12, 2015 at 19:17
• Are you able to prove that the perpendicular bisector of $T_1 T_2$ goes through the midpoint of a side of $ABC$? Oct 12, 2015 at 19:21
• You may also use this result: en.wikipedia.org/wiki/Simson_line Oct 12, 2015 at 19:21
• Yes i can prove that, but i don't know what i have to do with the "Simson line" Oct 12, 2015 at 19:36
• I just provided a proof, let me know if it is clear enough to you. Oct 12, 2015 at 19:53 Let $G,H,I\equiv T_1$ the feet of the altitudes of $ABC$. Since the circumcircle of the median triangle $DEF$ is the nine-point-circle of $ABC$, $D,E,F,G,H,I$ are concyclic. Since $\widehat{BT_3 C}=\widehat{CIB}=\frac{\pi}{2}$, $BCT_3 T_1$ is a cyclic quadrilateral and the perpendicular bisector of $T_1 T_3$ goes through $D$. Let $T_4$ be the midpoint of $T_2 T_3$. By Thales' theorem, the perpendicular bisector of $T_2 T_3$ goes through $F$. Now the angle between the $FT_4$ and the $JD$ lines equals the angle between the $CT_3$ and $T_1 T_3$ lines, by switching to perpendiculars. By exploiting the ciclicity of $BCT_3 T_1$ we have that this angle equals $\widehat{CBA}=\widehat{FED}$. If $J$ is the intersection between $FT_4$ and the perpendicular bisector of $T_1 T_3$, $J$ is the circumcenter of $T_1 T_2 T_3$ and $\widehat{FJD}=\widehat{FED}$. It follows that $J$ belongs to the nine-point-circle of $ABC$, as wanted.