If AM=BM, then point M is a middle point of side AB.
1. Consider triangle AML. You know that AM=ML, then this triangle is isosceles and AL is its base. The angles adjacent to the base of isosceles triangle are conruent, this means that [tex]\angle MAK\cong \angle AML.[/tex]
2. Consider lines ML and AC. The angle bisector AL is transversal. Since alternate interior angles [tex]\angle MAK\cong \angle AML,[/tex] you have that lines ML and AC are parallel. This means that ML is a middle line of triangle and 2ML=AC. Also you know that AC=2AL. This gives you that ML=AL. Now ML=AL and ML=AM gives you that triangle AML is equilateral.
3. In equilateral triangle AML all angles are congruent and have measures 60°. Thus, m∠AML=m∠MLA=m∠LAM=60°.
4. AL is angle bisector, then m∠MAL=m∠LAC=60° and m∠BAC=m∠MAL+m∠LAC=120°.
5. Consider ΔBML, it is isosceles, because BM=ML and m∠BML=180°-m∠AML=180°-60°=120°. Then,
[tex]m\angle MBL=m\angle MLB=\dfrac{180^{\circ}-120^{\circ}}{2}=30^{\circ}.[/tex]
6. Consider triangle ABC. In this triangle m∠A=120°, m∠B=30°, then
m∠C=180°-m∠A-m∠B=180°-120°-30°=30°.
Answer: m∠A=120°, m∠B=m∠C=30°.
