Constructing perpendiculars and triangles

Problem 1. Perpendicular to line a at its point O: a circle centred at O meets a at A and B; circles centred at A and B with radius AB meet at C; line OC is perpendicular to a. Justification: in △AOC and △BOC, AO = BO, AC = BC and CO is common, so by SSS ∠AOC = ∠BOC; they are adjacent and equal, so each is 90°. Problem 2. Perpendicular from a point O not on a: a circle centred at O meeting a gives A and B; circles of the same radius centred at A and B also meet at O₁; OO₁ ⊥ a. Theorem: through a point not on a line exactly one perpendicular to it can be drawn. Problem 3. Bisecting segment AB: circles of radius AB centred at A and B meet at C and C₁; the point where line CC₁ meets AB is the midpoint of AB.