Question 1. It is proved that the vertical line passing through O 1 and 02 respectively as AD intersects with AD at E and F.

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