Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 3 - Review Exercises - Page 166: 22



Work Step by Step

Step one : since the triangle ABC is an isosceles triangle so that the base angles are congruent which means $ 180=m\angle B + \angle A+m \angle B $ By substitution $ 180=50+2\angle A $ So $ \angle A=\angle B= 65 $. Step two : $ \angle C= \angle BCD + \angle DCA = 65^{\circ} and \angle A= \angle BAD + \angle DAC = 65^{\circ} $ since angle A equal to Angle C and given Angle BCD equal to angle DAC therefore, we conclude that $ \angle DCA= \angle BAD $ . Step three : by constructing an exterior angle of a triangle ADC and we know that the exterior angle of a triangle equal the sum measures of the two non adjacent interior angles. The exterior angle of triangle ADC is $ 115 + \angle BCD =\angle ADC + \angle DAC $ Since its given that angle BCD is congruent to angle DAC , so by substitution $\angle ADC = 115^{\circ} $.
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