Ceva's Theorem in $ BEK \Rightarrow AE, BG $ and $ CK $ are concurrent $ \iff \frac {sin_{BKC}} {sin_{CKE}} . \frac {sin_{KEA}} {sin_{AEB}} . \frac {sin_{EBG}} {sin_{GBK}} = 1 $. $ \iff \frac {sin_{\alpha}} {sin_{2\alpha}} . \frac {sin_{4\alpha}} {sin_{\alpha}} . \frac {sin_{2\alpha}} {sin_{4\alpha}} = 1 $. Which is clearly true.

Attachments:

Note that $AG$, $CK$, and $BE$ are the angle bisectors of triangle $AGC$, so they must be concurrent at the incenter of this triangle.