In quadrilateral ABCD the bisector of ∠A and the bisector of ∠B meet at the point P. If ∠C = 100° and ∠D = 70°, what is the measure of ∠APB?

The four angles of a quadrilateral add to 360°, so ∠A + ∠B = 360° − 100° − 70° = 190°. In triangle APB the angles at A and B are the halves ∠A/2 and ∠B/2, so ∠APB = 180° − ½(∠A + ∠B) = 180° − 95° = 85°. Rearranged, this says ∠APB = ½(∠C + ∠D) — a neat result, and the reason the last option is wrong: only the SUM of ∠A and ∠B is needed, and that is already fixed. The trap 95° stops at the half-sum ½(∠A + ∠B), which is the supplement of the answer, and 170° is ∠C + ∠D with the halving forgotten.