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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...

  • $\begingroup$ How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information $\endgroup$ – user10354138 Dec 10 '18 at 4:27
  • $\begingroup$ $G$ 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. $\endgroup$ – Derek Adams Dec 10 '18 at 5:52
  • $\begingroup$ @DerekAdams, did you understand my answer? $\endgroup$ – 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.



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