Bisecting an angle

To construct an eight-petalled figure, the supporting lines must divide the space around the centre into eight EQUAL angles. Since 360° is divided into 8 parts, each angle is 360° ÷ 8 = 45°.

So how do we construct a 45° angle with only a ruler and compass? We already know how to make 90°. If we can divide that into two equal parts — BISECT it — we get 45°. So let us find a general way to bisect ANY angle.

Take an angle ∠XOY. The idea is to build two CONGRUENT triangles, ∆OBC and ∆OAC, sharing the line OC. If they are congruent, then ∠BOC = ∠AOC, and OC bisects the angle.

How do we get such triangles? Mark A on one arm and B on the other with OA = OB — a single arc from O does that. Then choose C so that AC = BC. Now OA = OB, AC = BC, and OC is common to both triangles, so by the SSS condition ∆OBC ≅ ∆OAC.

That gives the steps. Step 1: mark points A and B such that OA = OB. Step 2: choosing any sufficiently long radius, cut arcs from A and from B, keeping the radius the same, and mark their intersection C. Step 3: OC bisects ∠AOB.

So a 45° angle is built by constructing 90° and then bisecting it. Bisect again for 22.5°, and again for 11.25°.

Two questions worth exploring by construction. In Step 2, if the equal arcs are drawn on the OTHER side, does OC still bisect the angle? (It does, by the same congruence argument.) And which angles can be built this way — could you construct 65.5°? Bisection halves, so from 90° you can reach 45, 22.5, 11.25 and so on, but not every angle you might name.