Question

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.

Answer 1

See explanation

Explanation:

If MA is tangent to the circle k(O), then radius OA is perpendicular to segment MA.

If MB is tangent to the circle k(O), then radius OB is perpendicular to segment MB.

Consider two right triangles MOA and MOB. In these triangles:

• MO is common hypotenuse;
• ∠OAM=∠OBM=90°, because MA⊥OA, MB⊥OB;
• OA=OB as radii of the circle k(O).

Thus, triangles MOA and MOB are congruent by HL theorem. So

∠AMO=∠BMO.

If point C is symmetric to point O with respect to point B, then OC⊥MB. Consider two right triangles MOB and MCB. In these triangles:

• MB is common leg;
• ∠OBM=∠CBM=90°, because OC⊥MB;
• OB=BC, because point C is symmetric to point O.

Thus, triangles MOB and MCB are congruent by HL theorem. So

∠BMO=∠BMC.

Hence,

∠AMC=∠AMO+∠BMO+∠BMC=3∠BMC.