# proving Thales’ theorem with vectors

Let the radius of the circle be $r$ and $AB$ a diameter of the circle. We make the dot product calculation

 $\overrightarrow{PA}\cdot\overrightarrow{PB}=(\overrightarrow{PO}+% \overrightarrow{OA})\cdot(\overrightarrow{PO}+\overrightarrow{OB})=(% \overrightarrow{PO}+\overrightarrow{OA})\cdot(\overrightarrow{PO}-% \overrightarrow{OA})=\overrightarrow{PO}\cdot\overrightarrow{PO}-% \overrightarrow{OA}\cdot\overrightarrow{OA}=r^{2}-r^{2}=0.$

The result shows that  $\overrightarrow{PA}\perp\overrightarrow{PB}$, i.e. the circumferential angle $APB$ of the half-circle is a right angle.

 Title proving Thales’ theorem with vectors Canonical name ProvingThalesTheoremWithVectors Date of creation 2013-03-22 17:47:45 Last modified on 2013-03-22 17:47:45 Owner pahio (2872) Last modified by pahio (2872) Numerical id 5 Author pahio (2872) Entry type Proof Classification msc 51F20 Classification msc 51F20 Synonym vector proof of Thales’ theorem Related topic ThalesTheorem Related topic SumOfVectors Related topic DifferenceOfVectors Related topic ScalarSquare Related topic ParallelogramPrinciple