Concyclicity (Proposed by Dao Thanh Oai)

$B$, $D$, $C$ are collinear. $A$ is another point on the plane. Draw $AB$, $AD$, $AC$, as in the figure. Rotate $AB$, $AD$, $AC$ from $B$, $D$, $C$ (couter clockwise or clockwise) the same angle. $\angle{EBA} = \angle{EDA} = \angle{FCA}$. $BE$ meets $CF$ in $F$; $DE$ meets $CF$ in $G$; $BE$ meets $DE$ in $E$.

Prove that $AEFG$ is cyclic.

Proof:

$\angle{EBA} = \angle{EDA} = \angle{FCA}$. The chord $EA$ is subtended by both angles $\angle{EBA}$, $\angle{EDA}$, so $ABDE$ is cyclic. Also, chord $AF$ is subtended by both angles $\angle{FBA}$, $\angle{FCA}$, so $ABCF$ is cyclic. Chord $AD$ is subtended by both angles $\angle{AED}$, $\angle{ABD}$. As $ABDE$ is cyclic $\angle{AED} = \angle{ABD}$. Chord $AC$ is subtended by both angles $\angle{AFC}$, $\angle{ABC}$. As $ABCF$ is cyclic $\angle{AFC} = \angle{ABC}$. Hence $\angle{AEG} = \angle{AFG}$. Indeed because the chord $AG$ is subtended by both angles $\angle{AEG}$, $\angle{AFG}$. Of course, this implies $AEFG$ is cyclic.

$Q.E.D.$