Problem:
(*) Let ABCD be a quadrilateral inscribed in circle ω with AC⊥BD. Let E and F be the reflections of D over lines BA and BC, respectively, and let P be the intersection of lines BD and EF. Suppose that the circumcircle of △EPD meets ω at D and Q, and the circumcircle of △FPD meets ω at D and R. Show that EQ=FR.
(c) 2018, Mathematical Association of America.
Solution:
First solution. Let X and Y be the feet of the perpendiculars from D to lines BA and BC, respectively, and let Z be the intersection of lines BD and AC. By Simson's theorem, the points X,Y,Z are collinear. A homothety with ratio 2 about D maps X,Y,Z to E,F,P′, respectively, where P′ is the orthocenter of △ABC. Hence, P′ lies on line EF as well as line BD, so P′=P.
Suppose now we extend ray CP to meet ω again at Q′. Then line BA is the perpendicular bisector of both PQ′​ and DE; consequently, PQ′ED is an isosceles trapezoid. In particular, it is cyclic, and so Q′=Q. In the same way, R is the second intersection of ray AP with ω.
Now, because of the two isosceles trapezoids we have found, we conclude
EQ=PD=FR
as desired.
Second solution. Here is a solution which does not identify the point P at all. We know that BE=BD=BF, by construction.
Claim 1. The points B,Q,E are collinear. Similarly the points B,R,F are collinear.
Proof. Work with directed angles modulo 180∘. Let Q′ be the intersection of line BE with circle ω (distinct from B ). Let α=∠DEB=∠BDE and β=∠BFD=∠FDB.
We know that BE=BD=BF, so B is the circumcenter of △DEF. Thus, ∠DEP=∠DEF=90∘−β. Then
