# Proving a quadrilateral is cyclic I am given that, for $$JG$$ an exterior angle bisector of $$\angle CGF$$ parallel to the angle bisector of $$\angle FHE$$, prove that $$CDEF$$ is a cyclic quadrilateral. I can prove that if the quadrilateral is cyclic, then the angle bisectors are parallel, but the reverse direction is proving troublesome...

• How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information – user10354138 Dec 10 '18 at 4:27
• $G$ is the intersection of the extension of sides $DC$ and $FE$, and $H$ is constructed similarly with sides $CF$ and $DE$. $J$ is an arbitrary point on the exterior angle bisector of $\angle DGE$. All that is known of $D,E,F,C$ is that they lie on the same circle. – Derek Adams Dec 10 '18 at 5:52
• @DerekAdams, did you understand my answer? – Anubhab Ghosal Dec 10 '18 at 13:00 Let $$\angle CGF=2\alpha$$, $$\angle FHE=2\beta$$ and $$\angle KCH=\gamma$$, as shown in the diagram.
$$\therefore\angle CKH=\pi-\gamma-\beta$$ and $$\angle JGK=\frac{\pi}{2}-\alpha$$ .
$$\angle CDE=\pi-\gamma-2\beta$$ and $$\angle CFG=\gamma-2\alpha$$
Thus, $$\angle CKH=\angle JGK$$ if and only if $$\angle CDE=\angle CFG$$, whence, exterior angle bisector of $$\angle CGF$$ is parallel to the angle bisector of $$\angle FHE$$, if and only if $$CDEF$$ is a cyclic quadrilateral.
$$\blacksquare$$