Hello, I'm studying with a group of friends going through a geometry study guide but we don't know any of the answers because we can't check our answers anywhere. If anyone can help or give a hint, it would be greatly appreciated! So far we think that angles AKE and AKF are 90 degrees but we don't know if that's possible. The line AM is a transversal to the two given parallel lines so that's where we reached that conclusion. Then BM is congruent to MC too.
Again, thanks a lot!
The triangles $AFK$ and $ABM$ are similar since they have equal angles. (The fact that $BC$ is parallel to $FE$ implies that the angles $AFK =ABM$) thus $BM/FK = MA/KA$. The triangles $AMC$ and $AKE$ are also similar so $MC/KE=MA/KA$. Thus $MC/KE =BM/FK$ since $BM=MC$, $KE=FK$.