Given any triangle $\triangle ABC$, let denote with $D$, $E$ and $F$ the midpoints of the three sides, and draw the three circles with centers in $D,E,F$ and passing by $A,B,C$, respectively.
These three circles determine other three points $G,H,I$ in correspondence of their intersections (other than $A,B,C$).
My conjecture (likely pretty obvious, like this one) is that
The points $D,E,F,G,H,I$ are always concyclic.
In order to prove this, I think one must show that, e.g. the points $D,F,E,H$ lie on a regular trapezoid, but my attempts so far yield to a really muddled reasoning.
Any suggestion how to sketch a simple proof of such conjecture?
Thanks for your help!
Best Answer
Triangle FHE and triangle FCE are congruent. (observe circles centered at E and F)
Triangle FCE and triangle DEF are congruent. ( $\because$ D, E, F are midpoints)
Therefore, Triangle FHE and triangle DEF are congruent.
Let S be the circle that D, E, F lie on. Then S is symmetric about the perpendicular bisector of line segment EF. ( $\because$ the center of S lies on the bisector)
The union of triangle FHE and triangle DEF is also symmetric about the perpendicular bisector of line segment EF. ( $\because$ the two triangles are congruent)
Since D is on the circle S, H is also on S.
Hence the points D, F, E, H are concyclic.
Note:
The name of the circle S is nine-point circle.