Proof of the central angle theorem

We prove that the central angle is twice the inscribed angle subtending the same arc. The key idea is to draw the diameter from the vertex of the inscribed angle and to exploit the isosceles triangles that form with the radii.

The diameter $CD$ divides the inscribed angle $\widehat{ACB}$ and creates two isosceles triangles $COA$ and $COB$.

Proof — Central angle

1. We draw the diameter $CD$ through $C$ and the centre $O$. The diameter divides the inscribed angle $\widehat{ACB}$ into two parts.
2. The triangles $COA$ and $COB$ are isosceles because $CO\cong OA\cong OB$ (radii).
3. In an isosceles triangle the base angles are congruent: $\widehat{OCA}\cong\widehat{OAC}, \qquad \widehat{OCB}\cong\widehat{OBC}.$
4. Now consider $\widehat{DOA}$: it is the exterior angle of triangle $COA$, hence: $\widehat{DOA} = \widehat{OCA}+\widehat{OAC} = 2\,\widehat{OCA}.$
5. Similarly: $\widehat{DOB} = 2\,\widehat{OCB}$.
6. Adding the two relations: $\widehat{AOB} = \widehat{DOA}+\widehat{DOB} = 2(\widehat{OCA}+\widehat{OCB}) = 2\,\widehat{ACB}.$

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Topics: Euclidean circle
Concepts: Central angle · Inscribed angle
Skills: Proving · Synthetic geometry