In triangle ABC the angle at A measures 100°. The perpendicular bisector of the side BC cuts the side AB at the point P, and ∠ACP = 20°. What is the measure of ∠ABC?

P lies on the perpendicular bisector of BC, so PB = PC and triangle PBC is isosceles: ∠PCB = ∠PBC. Since P is on the side AB, ∠PBC is simply ∠ABC, call it β, so ∠PCB = β as well. Then ∠ACB = ∠ACP + ∠PCB = 20° + β, and the angle sum gives 100 + β + 20 + β = 180, so β = 30°. Everything checks: ∠C = 50°, triangle PBC has angles 30°, 30° and 120°, and triangle APC has 100°, 20° and 60°. The trap 40° halves the leftover 80° as if the triangle were isosceles at A, 20° copies the given angle, and 50° answers ∠ACB instead.