In a triangle ABC, through circumcenter O we draw three lines parallel to the triangle sides. Call the intersection of lines with the triangle as D,E,F,G,H,K as on the image. Connect the neighboring of these which do not lie on the same side, as on the image. Prove that perpendicular bisectors on these new segments (KD, EG, FH) are concurrent.
I found four properties but cannot progress further. (check image for reference)
- A'B'C' has circumcircle also centered at O with radius 1/2 of the (ABC) one. Thus homothety from O with factor 2 into larger circle.
- B'C' C'A' A'B' are also parallel with respective sides of the triangle and at half the distance between the respective side and it's parallel that is passing through O.
- If you extend KD, EG and HF they form a new triangle A1, B1, C1. Surprisingly these A1, B1, C1 are collinear with CC'O AA'O and BB'O. (but don't know how to prove it)
- Call the intersection that needs to be proven as I. Then A1B'IC' are concyclic. Same for B1 and C1.
Comment: Triangle ABC is isosceles and lines parallel with sides pass O. As can be seen in figure , GE and DK are two chords of a circle and their perpendicular bisectors pass through I , the center of this circle. Since AB=AC the parallel line with side BC will have it's midpoint on O. FH is parallel with this line so the perpendicular bisector of FH is also that of GK and BC, hence it is the altitude of ABC from A and it passes through I.