This is question from math olympiad, can anyone help me to solve this question?

Question

Consider 5 points, A, B, C, D, and E such that ABCD is a parallelogram and BCED is a cyclic quadrilateral. Let l be a line passing through A. Suppose that l intersects the interior of the segment DC at F and intersects line BC at G. Suppose also that EF = EG = EC. Prove that l is the bisector of angle DAB.

Answer 1

The problem will be solved if we prove that EB=ED.
In this case, the triangle EBD is isosceles with equal angles DBE and BDE. The angle DCE=DBE because it is inscribed on the same side of the chord DE. The angles BCE and BDE are inscribed on the opposite sides of the chord BE. Hence, BDE = 180 - BCE = GCE and GCE=DCE. Equality of angles GCE and DCE means that the triangles CGE and FCE are equal. Consequently, CF=CG and the triangle FCG is isosceles. Since triangles FCG and FAD are similar, then the triangle FAD is also isosceles: DA=DF. The angle FAD is equal to:

FAD = (π - ADF)/2 = DAB/2,

which proves the statement.


Now, we need to show that EB=ED.
Let's drop the perpendicular from E to the line BD. The perpendicular foot is M. EB=ED if and only if BM=MD. Let's drop perpendiculars form point E to the line BG (foot Q) and to the line DC (foot P). The triangle CBD is inscribed, and the point E lies on the circumcircle. Hence, according to the Simson theorem, the points Q, P, and M are collinear, i.e. they belong to the same line (the Simson line). According to the Menelaus theorem, collinear points lying on (continuations of) triangle sides satisfy the condition

(DP/PC) * (CQ/QB) * (BM/MD) = 1.

Let's calculate the ratio DP/PC. The triangle CEF is isosceles, and EP is the altitude, so that PC=FP=CF/2. Since DP = DF + FP=DF+CF/2, then DP/CP = (DF + CF/2) / (CF/2) = 2*DF/CF + 1. DF/CF=AD/CG, because triangles FAD and FCG are similar, and AD=BC, because ABCD is parallelogram. Then DP/CP = 2*BC/CG + 1 = (2*BC+CG)/CG. The triangle CGE is isosceles, and EQ is the altitude, so that CG=2*CQ. Then

DP/CP = (BC+CQ)/CQ = BQ/CQ.

Substituting this equation in the Menelaus condition, we see that BM/MD=1. This means that

EB=ED,

which finishes the prove.