A, B, C and D lie on a circle in that order, and the chords AC and BD cross at the point P inside the circle. If ∠CAD = 35° and ∠ADB = 25°, what is the measure of ∠APB?

P lies on AC, so ∠PAD = ∠CAD = 35°; P lies on BD, so ∠PDA = ∠ADB = 25°. In triangle APD that leaves ∠APD = 180° − 35° − 25° = 120°. Since B, P and D are collinear, ∠APB is the exterior angle of triangle APD at P, so ∠APB = 35° + 25° = 60° (equivalently 180° − 120°). The trap 120° is ∠APD, the angle on the other side of line BD; 70° doubles 35° as though P were the centre.