Make auxiliary lines and connect AM, AN an. As we all know, AME is a right triangle with ∠AME as the right angle. As long as MF is proved to be the center line of the side AE, then the problem is proved.

According to the meANing of the question: AM=AN and because m is the midpoint of arc AC, we can know that arc MC= arc AM= arc an. Then because of the arc length, etc.:

∠MAC=∠AMN. So AF=MF.

And ∠ MAC+∠ MEA = 90, ∠ AMN+∠ FME = 90. Available: ∠MEA=∠FME. Then MF=FE.

Available: AF=MF=FE. that is

:AF=EF .