In △O 1EM and △O2FM,
∫o 1M = O2M (known) ≈ emo1= ∠ fmo2 ∠ mo1e = ∠ mo2f.
∴△o 1em≌△o2fm ∴o 1e=o2f
∴AB=CD (the isocenter in the same circle or the same circle is equal to the opposite chord)
EM=FM
∫o 1a = o2d (equal circle radius) o1e = o2f ∴ rt △ o1AE ≌ rt △ o2df (HL)
∴AE=FD
∴AM=MD
Question 2
Proof 1. Connection, OM
∫∠NOF = 2∠NCF (the central angle of the same arc is equal to 2 times the circumferential angle)
∠FOM=2∠FCM (same as above)
Arc NF= arc FM
∴∠NOF=∠FOM (the central angles of equal arcs in the same circle are equal)
∴∠NCF=∠FCM, that is, CF shares ∠NCM.
2.∠∠NCF =∠FCM (certification)
∠NOF=2∠NCF ∠MCN=2∠NCF
∴∠NOF=∠MCN
∴OF‖CM
∴OF⊥AB
∴ Arc AF= Arc FB (vertical diameter theorem)
∴ arc AM= arc NB