Question

PLEASE HELP ASAP!!!!!!!! Thank you!
Lines MA and MB tangent circle k(O) at A and B. Point C is symmetric to point O with respect to point B . Prove: m∠AMC=3m∠BMC.
PLEASE PROVE THOROUGHLY, or try your best :)

Answer 1

Answer:

Since OB is a radius and MB is a tangent, OB and MB are perpendicular.

So ∠MBO = ∠MBC = 90°.  Also, OB=CB by definition of C, and MB is common, so triangles MBO and MBC are congruent.  It follows that ∠BMC = ∠BMO.

Also, triangles MBO and MAO are congruent (common side MO, OB=OA are radii, and MA=MB as these are tangents from a common point).  So ∠BMO = ∠AMO.

Thus ∠BMC = ∠BMO = ∠AMO.

Finally,

                   3 ∠BMC = ∠BMC + ∠BMO + ∠AMO = ∠AMC.