It is proved that the diameter of the circle passing through B is BA', which connects Ca'; Then ∠A'CB is a fillet with a circular diameter, then ∠A'CB=90? ; ∠A=∠A '

=∠CBN； In RT△A'BC, ∠A'BC+∠A'=∠A'BC+∠CBN=∠A'BN=90? , that is, MN⊥A'B, so MN is the tangent of circle O.