Exercise 339Geometry › Plane geometryHigh schoolAdded 31 Oct 2026

Thales' theorem

#circles#angles

Statement

Let $[AB]$ be a diameter of a circle with centre $O$ , and let $M$ be a point of the circle different from $A$ and $B$ . Prove that the angle $\widehat{AMB}$ is a right angle.

Hint

Triangles $OAM$ and $OBM$ are isosceles.

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Solution

Since $OA=OM=OB$ , the triangles $OAM$ and $OBM$ are isosceles at $O$ . Let $\alpha=\widehat{OAM}=\widehat{OMA}$ and $\beta=\widehat{OBM}=\widehat{OMB}$ .

The angle $\widehat{AMB}$ equals $\widehat{OMA}+\widehat{OMB}=\alpha+\beta$ .

The angles of triangle $ABM$ are $\alpha$ (at $A$ ), $\beta$ (at $B$ ) and $\alpha+\beta$ (at $M$ ). Their sum is $180^\circ$ , so $2(\alpha+\beta)=180^\circ$ , hence $\widehat{AMB}=\alpha+\beta=90^\circ$ .

Thales' theorem

Statement

Solution

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