The Perfect Eye: Perpendicular Bisectors

Find the exact centres for a symmetrical eye, prove with congruence that they give the perpendicular bisector, and construct it in three steps.

9 micro-lessons

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The Eye's Hidden Centres

For a symmetrical eye, both arcs need centres equally far from X and Y.

An eye is made of an upper arc and a lower arc that meet at X and Y. The line XY 'supports' the drawing, though it isn't part of it.

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A two-step congruence proof that works for any length and radius.

Problem. Show that AB is always the perpendicular bisector of XY, whatever the length of XY and the radius used. (Textbook pp. 137–138)

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Put the steps of the perpendicular-bisector proof in order.

Drag the steps into the right order, first step at the top.

Drag to reorder

Explain why every equidistant point lies on the bisector.

Justify the statement

Textbook p. 139. Justify this statement using the facts we have established:

Any point that has the same distance from X and Y lies on the perpendicular bisector of XY.

Think of A and B.

Put the two facts together.

Arcs above, arcs below, join the crossings.

Using only an unmarked ruler and a compass, given a segment XY:

With a fixed radius (more than half of XY), draw arcs from X and from Y above XY. Name their meeting point A.
With the same radius, draw arcs from X and from Y below XY. Name their meeting point B.
Join AB. It is the perpendicular bisector.

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Bisecting a 6-unit segment XY, with every arc shown.

Problem. Construct the perpendicular bisector of a segment XY of length 6 cm.

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Explore which radii must be equal and which need not.

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Must the arcs above XY and the arcs below XY use the same radius? (Textbook p. 140, Q1)