see http://www.mathlinks.ro/Forum/viewtopic.php?t=208506 in my solution ,I obtain ∠PAD=∠QAF ∠PAF=∠QAD (different letters) thus our problem is solved

It’s not a big deal to show that C1 lies on (ABO), ÐAC1B=2ÐA, then C1C is its angle bisector: the triangles A’CA and B’BC are isosceles, then ÐAC1B=ÐA+ÐA’AC+ÐB’BC=2ÐA. Further, draw perpendiculars from A and B on CC0, call M and N their feet on it (obviously AM=BN, C0 being the middle of AB) and the perpendiculars from C on AA’ and BB’, call P and Q their feet onto these lines respectively. By symmetry, the triangles ACM and ACP are congruent, therefore AM=CP, similarly BN=CQ ( Here, the sign Ð means "angle"). Best regards, sunken rock

In may solution i made a parallelogram ACBC2 (C2 is a point) and i have a point S so <CSA=<AC2C so quadritetral AC2SC is cyclic,therofore by angle chassing we have <CBS=<ACB and we have <CSA=<CSC1=<CBC1 ,therofore C1BSC is cyclic,therofore <CBS=<CC1S = <ACB, therofore we have <AC1C= 180-<ACB and therofore we have <ACC1=<BCB1, as desired.

Dear Mathlinkers, just to remind that CC1 is the C-symmedian of ABC. Perhaps another approach without angle is possible. Sincerely Jean-Louis