Let $(ABCD)$ be a non-cyclical, convex quadrilateral, with no parallel sides. The lines $(AB)$ and $(CD)$ meet in $(E)$. Let $(M \neq E)$ be the intersection of the circumcircles of $(ADE)$ and $(BCE)$.
The internal angle bisectors of $(ABCD)$ form a convex, cyclical quadrilateral with circumcenter $(I)$. The external angle bisectors of $(ABCD)$ form a convex, cyclical quadrilateral with circumcenter $(J)$.
Show that $(I,J,M)$ are colinear.
Solution: Let $ W $ be the intersection of the external bisector of $ \angle A, $ $ \angle B $ and define $ X, $ $ Y, $ $ Z $ similarly. Let $ F $ be the intersection of $ AD, $ $ BC $ and let $ \Phi_M $ be the inversion with center $ M, $ power $ MA $ $ \cdot $ $ MC $ $ = $ $ MB $ $ \cdot $ $ MD $ $ = $ $ ME $ $ \cdot $ $ MF $ followed by reflection on the bisector $ \tau $ of $ \angle AMC, $ $ \angle BMD, $ $ \angle EMF. $ Since $ \measuredangle AZE $ $ = $ $ \measuredangle FYC, $ so by conformal $ \Longrightarrow $ $ \Phi_M(Z) $ lies on $ \odot (CFY). $ Similarly, we can prove $ \Phi_M(Z) $ lies on $\odot (BFW), $ so $ \Phi_M(Z) $ is the Miquel point of the complete quadrilateral $ BWYC $ and $ \Phi_M(Z) $ $ \in $ $ \odot (J). $ Similarly, we can prove $ \Phi_M(W), $ $ \Phi_M(X), $ $ \Phi_M(Y) $ lie on $ \odot (J), $ so $ J $ $ \in $ $ \tau. $ Analogously, we can prove $ I $ $ \in $ $ \tau. $
Could someone solve this problem by elementary mathematics, please?
I also don't believe the above "solution" is totally correct. (Z prime lies on circumscribed circle of CFY)