Thales’ theorem is a classic result about circles and right triangles: if \(A\) and \(B\) are the endpoints of a diameter of a circle,
and \(P\) is any other point on the circle, then the angle \(\angle APB\) is a right angle. In words, a diameter always subtends
\(90^\circ\) at the circumference. This theorem is important because it creates a direct bridge between circle geometry and
right-triangle geometry: every point on the semicircle determines a right triangle with hypotenuse \(AB\).
The cleanest proof uses the inscribed angle theorem. First, define the circle’s center \(O\) as the midpoint of the diameter
\(AB\). By construction, \(OA=OB\) because both are radii, and the central angle \(\angle AOB\) intercepts the same arc \(AB\) as the inscribed
angle \(\angle APB\). The inscribed angle theorem states that an angle formed by two chords at the circle is half the measure of the central
angle that intercepts the same arc. Symbolically,
\[
\angle APB = \tfrac12 \angle AOB.
\]
When \(AB\) is a diameter, points \(A\), \(O\), and \(B\) lie on a straight line, so \(\angle AOB = 180^\circ\). Substituting gives
\[
\angle APB = \tfrac12 \cdot 180^\circ = 90^\circ,
\]
which is exactly Thales’ theorem. Notice the statement is independent of where \(P\) lies on the circle (as long as \(P\neq A,B\)):
the right angle is guaranteed because the intercepted arc is fixed by the diameter.
There is also an intuitive “equal radii” angle-chasing proof. Connect the center \(O\) to \(A\), \(B\), and \(P\). Because \(OA=OP\) and
\(OB=OP\), triangles \(OAP\) and \(OBP\) are isosceles, so they have equal base angles at \(A\) and \(P\), and at \(B\) and \(P\). These equalities
let you express \(\angle APB\) as a sum of two angles that together make a straight angle around the diameter, again forcing \(\angle APB\) to be
\(90^\circ\). This approach helps you see why the circle’s symmetry matters: all radii are the same length, so the “turning” at \(P\) becomes
rigidly determined.
In coordinate geometry, you can verify the right angle with vectors. If you compute the vectors \(\overrightarrow{PA}=A-P\) and
\(\overrightarrow{PB}=B-P\), then \(\angle APB\) is \(90^\circ\) exactly when their dot product is zero:
\[
(A-P)\cdot(B-P)=0.
\]
When \(P\) is constrained to the circle with diameter \(AB\), this dot product becomes (up to rounding) zero for every position of \(P\),
matching the geometric theorem.
Thales’ theorem is used in construction and proofs: to create a right angle through a segment \(AB\), draw the circle with diameter \(AB\) and
choose any point \(P\) on the semicircle; triangle \(APB\) will be right. It also appears naturally in the study of inscribed angles, cyclic
quadrilaterals, and many problems where recognizing a diameter unlocks a hidden \(90^\circ\) angle.
