In triangle ABC, AB = AC and the angle at A measures 40°. The side BC is produced beyond C to a point D such that CD = CA. What is the measure of angle ADC?

Since AB = AC, angle ACB = (180° − 40°)/2 = 70°. As B, C and D lie on one straight line, angle ACD = 180° − 70° = 110°. Triangle ACD is isosceles because CA = CD, so its two base angles are equal: angle CAD = angle ADC = (180° − 110°)/2 = 35°. Forgetting to take the supplement at C and using 70° in the isosceles step would give 55°.