Prove by vector method, the angle subtended on a semicircle is a right angle.

Answer: With O as the centre, \(\overline{AP} \cdot \overline{BP} = |\bar{p}|^2 - |\bar{a}|^2 = 0\), so \(\angle APB = 90^\circ\).

Let \(AB\) be a diameter of a circle with centre O and radius \(r\), and let P be any point on the semicircle. Take O as the origin and let \(\bar{a}\) and \(\bar{p}\) be the position vectors of A and P. Since O is the midpoint of AB, the position vector of B is \(-\bar{a}\).

\(|\bar{a}| = |\bar{p}| = r\) (radii).

\(\overline{AP} = \bar{p} - \bar{a}\) and \(\overline{BP} = \bar{p} - (-\bar{a}) = \bar{p} + \bar{a}\).

\(\overline{AP} \cdot \overline{BP} = (\bar{p} - \bar{a}) \cdot (\bar{p} + \bar{a}) = |\bar{p}|^2 - |\bar{a}|^2 = r^2 - r^2 = 0\).

Both vectors are non-zero, so \(\overline{AP} \perp \overline{BP}\), i.e. \(\angle APB = 90^\circ\). Hence the angle in a semicircle is a right angle.