Show that angle in a semi circle is a right angle.

Let AB be a diameter of a semi circle with centre ‘O’. Let P be a point on semi circle other than A, B. Let position vector of P be $\overline{r}$ and position vector of A be $\overline{a}.$

i.e., $\overline{OP}=\overline{r},\overline{OA}=\overline{a}\text{ then }\overline{OB}=-\overline{a}$

$\therefore \overline{PA}=\overline{OA}-\overline{OP}=\overline{a}-\overline{r}$

$\overline{PB}=\overline{OB}-\overline{OP}=-\overline{a}-\overline{r}=-(\overline{a}+\overline{r})$

Now $\overline{PA}\cdot \overline{PB}=(\overline{a}-\overline{r})\cdot \{-(\overline{a}+\overline{r}\left)\right\}$

$=-\left[(\overline{a}{)}^2-(\overline{r}{)}^2\right]={\overline{r}}^2-{\overline{a}}^2=|\overline{r}{|}^2-|\overline{a}{|}^2$

$=0(\because |\overline{r}|=|\overline{a}|=$radius of the circle)

$\therefore \overline{PA}\cdot \overline{PB}=0\Rightarrow \overline{PA}\perp \overline{PB}\therefore ⌊APB={90}^{\circ }$

$\therefore \text{ Angle in a semi circle is }{90}^{\circ }$