The eye, and the perpendicular bisector

In Grade 6 you constructed 'Eyes' — a shape made from a lower arc and an upper arc that look symmetrical. Then, you found the two centres A and B by spatial estimation. Now we can find them exactly.

The arcs define a line XY that SUPPORTS the drawing, though it is not part of the final figure. For the eye to be symmetrical — for XY's supporting line to be the line of symmetry — the upper and lower arcs must have the SAME radius. That is, AX = BX. And since AX = AY and BX = BY, all four are equal: AX = AY = BX = BY.

So how do we find such A and B? Simple: from X and from Y, draw arcs of the same radius above XY, and take A where they meet; do the same below XY and take B.

Now join A and B. Where does AB cross XY, and at what angle? It passes through the MIDPOINT of XY, and it is PERPENDICULAR to it.

Two names. A division of a line, or any geometrical object, into two identical parts is called BISECTION. A line that bisects a given line AND is perpendicular to it is the perpendicular bisector.

Does this always happen, for any length XY and any radius of arcs? Congruence answers it. Let O be where AB meets XY. In ∆ABX and ∆ABY we have AX = AY, BX = BY, and AB common — so by SSS they are congruent, giving ∠XAO = ∠YAO.

Now look at ∆AOX and ∆AOY. We have AX = AY, AO common, and the angle between them equal — so by SAS they are congruent. Therefore OX = OY, so O is the midpoint; and ∠AOX = ∠AOY, which together make a straight angle of 180°, so each is 90°. AB really is the perpendicular bisector of XY, always.