Editing Thales's theorem (section)

===Proof of the converse using linear algebra===
This proof utilizes two facts:
*two lines form a right angle if and only if the [[dot product]] of their directional [[vector (geometry)|vector]]s is zero, and
*the square of the length of a vector is given by the dot product of the vector with itself.
Let there be a right angle {{math|∠ ''ABC''}} and circle {{mvar|M}} with {{overline|AC|}} as a diameter.
Let M's center lie on the origin, for easier calculation.
Then we know
*{{math|1=''A'' = −''C''}}, because the circle centered at the origin has {{mvar|{{overline|AC}}}} as diameter, and
*{{math|1=(''A'' – ''B'') · (''B'' – ''C'') = 0}}, because {{math|∠ ''ABC''}} is a right angle.
It follows
:<math>\begin{align}
0 &= (A-B) \cdot (B-C) \\
  &= (A-B) \cdot (B+A) \\
  &= |A|^2 - |B|^2. \\[4pt]
  \therefore \ |A| &= |B|.
\end{align}</math>

This means that {{mvar|A}} and {{mvar|B}} are equidistant from the origin, i.e. from the center of {{mvar|M}}. Since {{mvar|A}} lies on {{mvar|M}}, so does {{mvar|B}}, and the circle {{mvar|M}} is therefore the triangle's circumcircle.

The above calculations in fact establish that both directions of Thales's theorem are valid in any [[inner product space]].