The four vertices of quadrilateral ABCD, named in order, lie on one circle. The sides AD and BC are produced beyond D and beyond C, and the two produced lines meet at the point E. If ∠ABC = 95° and ∠BAD = 70°, what is the measure of ∠AEB?

Opposite angles of a cyclic quadrilateral are supplementary, so ∠ADC = 180° − 95° = 85° and ∠BCD = 180° − 70° = 110°. Because D lies between A and E, ∠EDC = 180° − 85° = 95°, and because C lies between B and E, ∠ECD = 180° − 110° = 70°. In triangle EDC the third angle is ∠DEC = 180° − 95° − 70° = 15°, and that is the same angle as ∠AEB. The quickest route reaches the same value: in triangle EAB the angles at A and B are just ∠BAD = 70° and ∠ABC = 95°, so ∠AEB = 180° − 70° − 95° = 15°. The trap 25° differences the two given angles, 165° adds them and stops, and 10° halves the difference.