Grade 11 Euclidean Geometry • Theorem 1

Line from Centre to Chord

The Intuition (Plain English)

If you draw a line straight down from the centre of a circle so that it hits a chord at exactly 90 degrees (perpendicular), it will automatically slice that chord into two perfectly equal halves.

The Golden Rule: This theorem is essentially a two-way street. If the exam tells you the line is perpendicular, you know it bisects the chord. If the exam tells you the line cuts the chord in half, you know it must be hitting it at 90 degrees.

The Formal Proof

This is one of the 6 examinable proofs. You must be able to reproduce this exact layout in Paper 2.

[Insert GeoGebra Diagram:
Circle centre O, chord AB, line OM ⊥ AB]

Given: Circle with centre O. Chord AB and line OM such that OM ⊥ AB.

R.T.P: AM = MB

Construction: Draw radii OA and OB.

Statement	Reason
In ΔOMA and ΔOMB:
1. OA = OB	radii
2. M̂₁ = M̂₂ = 90°	given (OM ⊥ AB)
3. OM = OM	common side
∴ ΔOMA ≡ ΔOMB	RHS
∴ AM = MB	from congruency

How to apply this in an exam (The "Pythagoras Trap")

In Paper 2, they rarely test this theorem in isolation. They almost always combine it with the Theorem of Pythagoras to calculate lengths.

Typical Exam Question:
A circle has centre O and a radius of 13 cm. A chord PQ has a length of 24 cm. Calculate the shortest distance from the centre O to the chord PQ.

Step-by-Step Solution:

Let the shortest distance be the line OM. We know the shortest distance from a point to a line is a perpendicular drop. Therefore, OM ⊥ PQ.
Because OM ⊥ PQ, we know from our theorem that PM = MQ. (line from centre ⊥ to chord)
Therefore, PM = 24/2 = 12 cm.
Draw radius OP. We are given that OP = 13 cm.
In right-angled ΔOMP:
OM² + PM² = OP² (Pythagoras)
OM² + (12)² = (13)²
OM² + 144 = 169
OM² = 25
OM = 5 cm.

⚠️ Common Exam Traps

Forgetting the Acceptable Reason: If you write "radius bisects chord" or "perpendicular line", you will lose the reason mark. You MUST write exactly: line from centre ⊥ to chord (or line from centre to midpt of chord if using the converse).
Assuming the line goes through the centre: The theorem only applies if the line originates from the centre of the circle. If the question doesn't explicitly state "O is the centre", you cannot apply this theorem.