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Parallel lines, angle bisector and exterior angle of a triangle
Question:
In the figure below, the angle A is 58o and CD is parallel to AB. CE is an angle bisector which bisects the angle ACD and intesect AB at E. what is the degree measure of the angle CEB?
Solution:
since AB // CD (Given)
so angle A + angle ACD = 180o
because angle A and angle ACD are interior angles on the same side of the transversal CA, because AB // CD, so these two angles are supplementary
angle ACD = 180 - angle A = 180o - 58o = 122o
since CE is the angle bisector (Given)
so angle ACE = angle DCE = (1/2) angle ACD = (1/2)(122o) = 61o
since points A, E, B are on the same line, so the angle CEB is an exterior angle of triangle ACE
so angle CEB = angle A + angle ACE
because the exterior angle of a triangle is equal to the sum of two noadjacent interior angles
so angle CEB = angle A + angle ACE = 58o + 61o = 119o
therefore, the angle CEB is 119o