330. The Chord's Mirror · how the centre always bisects at right angles

Try to break it

Drag A and B around the circle. M is the chord's midpoint, and OM is always perpendicular to AB. Stretch the chord almost across a diameter or shrink it to a tiny segment — the right angle at M never tilts. The line from the centre to a chord's midpoint always meets the chord at 90°.

How you build it

Build a chord and the line from the centre to its midpoint, and see OM ⊥ AB.

• Point tool: mark O near the middle — the centre of the circle.
• Circle tool: click O as the centre, then click outward to set the radius.
• Point tool: mark A on the circle.
• Point tool: mark B at another spot on the circle — AB is the chord.
• Segment tool: join A to B to draw the chord.
• Midpoint tool: click A then B — it drops M exactly at the middle of the chord. No measuring needed.
• Segment tool: join O to M. The line from the centre to the midpoint of a chord always meets it at a right angle, so OM ⊥ AB (∠OMA = 90°).

The proof, step by step

Prove that the line from the centre to the midpoint of a chord is perpendicular to it.

• Consider triangles OMA and OMB.
• OA = OB (radii of the same circle) and AM = BM (M is the midpoint of AB). OM is common.
• Therefore, ΔOMA ≅ ΔOMB by the SSS congruence rule.
• ∠OMA = ∠OMB (c.p.c.t.). Since they form a linear pair, each angle is 90°.
• Hence, OM ⊥ AB.

Worked example

In a circle of radius 10 cm, a chord is at a distance of 6 cm from the centre. Find the length of the chord.

Let the chord be AB and centre O. OM ⊥ AB, OM = 6 cm, OA = 10 cm. In ΔOMA, AM² = OA² - OM² = 100 - 36 = 64. So AM = 8 cm. Since OM bisects AB, AB = 2 × AM = 16 cm.

• 8 cm
• 12 cm
• 16 cm — correct
• 20 cm