السلام عليكم ورحمة الله وبركاته
هذا حل جديد للسؤال الثالث من الجولة الرابعة التى انتهت واعلنت نتيجتها

Let P,M,N be the midpoints of BD,CF,CG respectively.ABCD is
a parallelogram, AC also passes through P . Easily seen P,M,N are colinear
(by homothety or they are in the same line parallel with L)
Since EF = EC
EM is perpendicular to CF
similarly EN is perpendicular to CG .
Thus MN is simson line of E w.r.t BCD but MN intersection BD = P
implies EP is perpendicular to BD
but P is midpoint of BD implies EB=ED
angle BDE = angle DBE--------(1
angle DBE = angle DCE , angle BDE = angle ACG --------(2
Since ACED is cyclic
(1) & (2)

angle DCE = angle ACG
implies angle CEF = angle CEG
implies EC is midline FG

implies triangle CFG is isoceles

implies angle DAG = angle AGC = (180 - angle FCG)/2 = angle BCD/2 = angle BAD/2