The Perpendicular from the Centre Bisects a Chord

AB is a chord. OM runs from the centre O and meets the chord at a right angle, landing at M.

M is always the exact midpoint: AM = MB, wherever you put the chord. The right angle is what makes this work — a line from the centre that meets the chord at any other angle does not cut it in half.

How to Use

Interactive Exploration

Drag either end of the chord around the circumference. The perpendicular swings round to follow it, and the two halves stay equal however long or short the chord gets.

Notice what happens as the chord moves closer to the centre: it gets longer and OM gets shorter, but the halves stay matched. The chord is kept clear of being a full diameter, where M would land on O and there would be no perpendicular left to draw.

Solving Problems

New positions sets a fresh question: one half keeps its length and the other is marked x. Tap the x on the diagram, or click the matching reveal, to check.

Hovering a reveal lights up the half it names, so AM and MB are never in doubt.

The Style tab changes the colours and the background paper.

Teaching Notes

• Lengths are shown to one decimal place in grid units. Unlike the angle theorems these do not come out whole — but the two halves are equal exactly, which is the point.
• Why it works: join O to A and O to B. Both are radii, so triangle OAB is isosceles, and the perpendicular from the apex of an isosceles triangle bisects the base.
• The converse is just as useful and worth asking about: if a line from the centre bisects a chord, must it be perpendicular to it?
• Watch for students assuming any line from the centre to a chord bisects it. Ask them what happens to a line from O that meets the chord somewhere other than M.