- parallel line, angle bisector and exterior angle of a triangle example 2
- Question:
- In the figure below, AB // CD. CE bisects the angle ACD and intersect AB at E. If the angle A is 126
^{o}, what is the degree measure of the angle CEB?

- Solution:
- since AB // CD (Given)
- so angle A + angle ACD = 180
^{o}(If two parallel lines are cut by a transversal, then consecutive interior angles are supplementary.) - angle ACD = 180
^{o}- angle A = 180^{o}- 126^{o}= 54^{o} - since CE bisects the angle (Given)
- so angle ACE = angle DCE = (1/2)angle ACD = (1/2) × 54
^{o}= 27^{o} - since AEB is a straight line, so angle CEB is an exterior angle of the triangle ACE
- angle CEB = angle A + angle ACE (exterior angle of a triangle is equal to the sum of two noadjacent angles.)
- angle CEB = angle A + angle ACE = 126
^{o}+ 27^{o}= 153^{o} - so the degree measure of the angle CEB is 153
^{o}

The rule applied to solve this problem: when two parallel lines AB and CD are cut by a transversal AC, these two angles on the same side of the transversal are consecutive interior angles, and the consecutive interior angles are supplementary.
The supplementary means the sum of two angles is 180^{o}. The property of an angle bisector is cutting that angle in half. An exterior angle of a triangle is equal to the sum of two noadjacent angles.