Circle theorems: chord bisection
- Thales. The easiest circle theorem.
- Inscribed angle. The angle subtending a major arc is constant.
- Cyclic quadrilateral - opposite angles sum to 180°. Plus minor arc.
- Tangent radius. Blindingly obvious, but a slippery proof.
- Alternate segment. Perhaps the most unintuitive.
- Chord bisection by radius. The relevant angle is 90°.
- Equal length tangents. For completeness.
Chord bisection by radius theorem
If a chord AB is bisected at C by a radius OD then ∠OCQ = 90°. This is a simple and intuitively obvious. The set up is:
The proof
In the diagram we aim to show that ΔOAC and ΔOBC are congruent, so have the same angles.
In ΔAOC and Δ:
- Since both are radii, |OA| =|OB|.
- OC is a common side.
- |AC| = |BC| as OQ bisects AB.
By the SSS theorem - which most allow you to take as axiomatic - we conclude that that the two triangles are congruent. In particular ∠ACO = ∠BCO. But it is also clear that ∠ACO + ∠BCO = 180° so we conclude that ∠BCO = 90°, as required.
Where next?
Finally we have a theorem about equal length tangents.