259. The Bisector's Promise · perpendicular to every chord it meets
The line from the centre to the midpoint of a chord is always perpendicular to the chord.
What this lesson covers
Try to break it
Drag A and B around the circle (keep AB shorter than a diameter). The line from O to the chord always hits AB exactly at its midpoint AND meets it at 90°. Drag A and B so AB passes through O — then the chord becomes a diameter and the midpoint becomes O itself.
How you build it
Draw a chord and join its midpoint to centre.
- Point tool: mark the centre O.
- Circle tool: click O, then drag out to set the radius.
- Point tool: mark A on the circle.
- Point tool: mark B on the circle.
- Segment tool: join A to B — the chord.
- Midpoint tool: click A then B — C is the exact midpoint of the chord.
- Segment tool: join O to C — the line from the centre to the midpoint. It meets the chord AB at a right angle.
The proof, step by step
Prove that the line from the centre to the midpoint of a chord is perpendicular to the chord.
- Join OA and OB to form two triangles, ΔOAC and ΔOBC.
- OA = OB (radii), OC = OC (common), and AC = BC (C bisects AB).
- By SSS congruence, ΔOAC ≅ ΔOBC, so ∠OCA = ∠OCB.
- ∠OCA and ∠OCB form a linear pair (sum to 180°), so each is 90°.
- Therefore, OC is perpendicular to AB. ∎
Worked example
In a circle of radius 13 cm, a chord is 10 cm long. What is the distance from the centre to the chord?
The perpendicular from the centre bisects the chord, so half the chord is 5 cm. Using Pythagoras in the right triangle formed by the radius, half-chord, and distance: distance² + 5² = 13² → distance² = 169 - 25 = 144 → distance = 12 cm.
- 5 cm
- 10 cm
- 12 cm — correct
- 13 cm