In triangle ABC the point D lies on the side AB, and the lengths satisfy CD = CB. If ∠ACD = 20° and ∠ABC = 70°, what is the measure of ∠BAC?

Because CD = CB, triangle CDB is isosceles and the angles opposite those equal sides are equal: ∠CDB = ∠CBD. Now ∠CBD is just ∠ABC = 70°, so ∠CDB = 70° and the remaining angle gives ∠BCD = 180° − 70° − 70° = 40°. The whole angle at C is ∠ACB = ∠ACD + ∠DCB = 20° + 40° = 60°, so ∠BAC = 180° − 70° − 60° = 50°. Check the small triangle too: ∠ADC = 180° − 70° = 110°, and 50 + 20 + 110 = 180. The trap 60° answers ∠ACB, 40° answers ∠BCD, and 70° copies the given angle at B straight across.