Question

In △ABC AL is an angle bisector (L∈ BC ). Point M∈ AB so that LM=AM=BM. Find the angles in △ABC, if AC=2AL.

Answer 1

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 $\angle MAK\cong \angle AML.$

2. Consider lines ML and AC. The angle bisector AL is transversal. Since alternate interior angles $\angle MAK\cong \angle AML,$ 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,

$m\angle MBL=m\angle MLB=\dfrac{180^{\circ}-120^{\circ}}{2}=30^{\circ}.$

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°.