In triangle ABC, angle A is 40 ° and angle ECB adjacent to angle ABC is 80 ° Prove that the bisector

In triangle ABC, angle A is 40 ° and angle ECB adjacent to angle ABC is 80 ° Prove that the bisector of angle ECB is parallel to line AB.

1) Since, in triangle ABC, angle A = 40 °, and angle BCE = 80 °, then the inner angle C of triangle ABC is equal to:

Angle C = 180 ° – 80 ° = 100 °;

2) the angle B of the triangle ABC is equal to:

Angle B = 180 ° – 40 ° – 100 ° = 140 ° – 100 ° = 40 °.

3) Sides BC = AC, triangle ABC is isosceles.

4) The bisector of the angle BCE divides the angle in half.

CK is a bisector.

Angle BCK = angle KCE = 80 ° / 2 = 40 °.

Hence, we find that the bisector of the angle BCE is parallel to the side AB.