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  $\overrightarrow{PA}⟂\overrightarrow{PB}$, 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