# Circle : Cyclic and Concyclic

Two circles intersect at $\displaystyle C$ and $\displaystyle D$. $\displaystyle ABCD$ is a cyclic quadrilateral in one circle. $\displaystyle BC$ produced meets the other circle at $\displaystyle E$. Te points $\displaystyle C, F, E$ and $\displaystyle D$ are concyclic. $\displaystyle AB$ produced meets $\displaystyle EF$ produced at $\displaystyle G$. Prove that $\displaystyle GFDA$ is a cyclic quadrilateral.
$\displaystyle \begin{array}{l}\text{Given }\ \text{ : }ABCD\text{ is a cyclic quadrilateral}\text{. }\\\text{ }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ABG,BCE\ \text{and }EFG\ \text{are straight lines}\text{.}\\\\\text{To Prove : }GFDA\text{ is a cyclic quadrilateral}\text{.}\\\\\text{Proof : }\ \ \ ABCD\text{ is a cyclic quadrilateral}\text{.}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \theta +\beta =180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Since }D,C,F,E\ \text{are concyclic,}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta =\varepsilon \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{But }\varepsilon +\gamma =\beta \ \text{(Sum of interior }\angle \text{s of }\Delta \text{ = exterior }\angle \text{)}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta +\gamma =\beta \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \theta +\delta +\gamma =180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \angle ADF+\angle AGF=180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore D,C,F,E\ \text{are concyclic}.\end{array}$