I’m Up and Down, and Round and Round · Lesson 4 of 8
Midpoints and Perpendicular Bisectors of Chords
“The shortest path from the centre to a chord lands exactly at its midpoint—show-off.”
• Prove that the line from the centre to the midpoint of a chord is perpendicular to the chord. • Prove the converse result. • Understand the relationship between midpoint and perpendicular bisector. • Apply congruence to chord problems. • Use the theorem in inscribed-triangle reasoning.
A chord has an important relationship with the centre of a circle. Suppose a chord is drawn and its midpoint is marked. If we join this midpoint to the centre of the circle, the joining line meets the chord at a right angle. In other words, the line from the centre to the midpoint of a chord is always perpendicular to that chord.
This relationship also works in the reverse direction. If a line is drawn from the centre of a circle perpendicular to a chord, it divides the chord into two equal parts. This makes the centre, the midpoint of the chord, and the right angle closely connected, and the property is very useful when solving problems involving chord lengths, distances from the centre, and geometric constructions.
The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
Proof Step by Step
Let M be the midpoint of chord AB and O the centre. Compare triangles OMA and OMB.
| Equality | Reason |
|---|---|
| OA=OB | Radii of the same circle |
| AM=BM | M is midpoint of AB |
| OM=OM | Common side |
Therefore ΔOMA≅ΔOMB by SSS. Hence ∠OMA=∠OMB. These two angles form a straight line, so together they add to 180°. Equal angles adding to 180° must each be 90°. Therefore OM⊥AB.
The Converse
The perpendicular from the centre of a circle to a chord bisects the chord.
Suppose OM⊥AB. Then ∠OMA and ∠OMB are both 90°. OA=OB because they are radii, and OM is common. Therefore the two right triangles are congruent by RHS, which gives AM=BM.
Why These Two Results Are So Useful
Together, the two theorems let us move freely between two pieces of information. If we know midpoint, we get perpendicularity. If we know perpendicularity from the centre, we get the midpoint.
Problem
An isosceles triangle ABC is inscribed in a circle with AB=AC. Explain why the altitude from A to BC passes through the centre.
- 1.AB=AC, so A lies on the perpendicular bisector of BC.
- 2.The centre O of the circle is equidistant from B and C, so O also lies on the perpendicular bisector of BC.
- 3.Therefore A and O lie on the same perpendicular bisector of BC.
- 4.The altitude from A to BC is this perpendicular bisector, so it passes through O.
Practice Problems
- Prove that a perpendicular from the centre to a chord bisects the chord.
- In a circle, O is the centre and M is the midpoint of chord AB. What can you say about ∠OMA?
- An isosceles triangle ABC with AB=AC is inscribed in a circle. Prove that the altitude from A to BC passes through the centre.
- Two parallel chords lie on opposite sides of a circle's centre. Explain how their midpoints can be located using perpendiculars from the centre.
Key Takeaways
• Centre-to-midpoint of a chord is perpendicular to the chord. • Perpendicular from the centre to a chord bisects it. • The two statements are converses. • Congruent right triangles provide the proof. • These results connect chord length, midpoint and distance from the centre.
Next, we compare chords using their perpendicular distance from the centre.