(BMO 2018 Q1) A quadrilateral $ABCD$ is inscribed in a circle $k$, where $AB > CD$ and $AB$ is not parallel to $CD$. Point $M$ is the intersection of the diagonals $AC$ and $BD$ and the perpendicular from $M$ to $AB$ intersects the segment $AB$ at the point $E$. If $EM$ bisects the angle $CED$, prove that $AB$ is a diameter of the circle $k$.
My solutions are different from the official solution in https://bmo2018.dms.rs/wp-content/uploads/2018/05/Solutions.pdf, I am not sure these solutions are correct. Any constructive comments on my solution will be much appreciated.
Solution-1: Let $O$ be the center of circle $k$, Let $AB ∩ DC = P$ and $AD ∩ BC = Q$. We have a completed cyclic quadrilateral ($ABCDPQ$). Let $OM ∩ PQ = N$. By the properties of completed cyclic quadrilateral, we have $O,M,N$ are collinear. By Brocard Theorem, $△PMQ$ is self-polar and $O$ is the orthocenter of $△PMQ$, this implied that $QM ⊥ PO = E'$ and $OM ⊥ PQ = N$, we are given $ME ⊥ AB$, and we have $OP ≡ AB$, $⇒ E'= E$, $⇒ O$ must on $AB$, and $AB$ is the diameter.
Solution-2: Let $AB ∩ DC = P$ and $AD ∩ BC = Q$. We have a completed cyclic quadrilateral ($ABCDPQ$). Apply “Cevians induces harmonic lemma” on $△QAB$ we have -1 = ($P,E;A,E$), by the converse of “inversion induces harmonic lemma”, we can conclude that AB is the diameter of the circle $k$.