Editing Thales's theorem (section)

===Second proof===
The theorem may also be proven using [[trigonometry]]: Let {{math|1=''O'' = (0, 0)}}, {{math|1=''A ''= (−1, 0)}}, and {{math|1=''C'' = (1, 0)}}. Then {{mvar|B}} is a point on the unit circle {{math|(cos ''θ'', sin ''θ'')}}. We will show that {{math|△''ABC''}} forms a right angle by proving that {{mvar|{{overline|AB}}}} and {{mvar|{{overline|BC}}}} are [[perpendicular]] — that is, the product of their [[slope]]s is equal to −1. We calculate the slopes for {{mvar|{{overline|AB}}}} and {{mvar|{{overline|BC}}}}:

:<math>\begin{align}
m_{AB} &= \frac{y_B - y_A}{x_B - x_A} = \frac{\sin \theta}{\cos \theta + 1} \\[2pt]
m_{BC} &= \frac{y_C - y_B}{x_C - x_B} = \frac{-\sin \theta}{-\cos \theta + 1}
\end{align}</math>

Then we show that their product equals −1:

: <math>\begin{align}
&m_{AB} \cdot m_{BC}\\[4pt]
= {} & \frac{\sin \theta}{\cos \theta + 1} \cdot \frac{-\sin \theta}{-\cos \theta + 1}\\[4pt]
= {} & \frac{-\sin ^2 \theta}{-\cos ^2 \theta +1}\\[4pt]
= {} & \frac{-\sin ^2 \theta}{\sin ^2 \theta}\\[4pt]
= {} & {-1}
\end{align}</math>

Note the use of the [[Pythagorean trigonometric identity]] <math>\sin^2 \theta + \cos^2 \theta = 1.</math>