Prove that angle in a semi-circle is a right angle.

Toolbox:
• To prove angle between any two lines is rightangle, prove that the dot product of the two vectors =0
• $\large(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\overrightarrow a|^2-|\overrightarrow b|^2$
Let O be origin, A and A' be points at equal distance from O and
are on positive and negative x-axis. Let AA' be diametre of
semi circle. Let P be a point on the semicircle.
$\Rightarrow |\overrightarrow {OA}|=|\overrightarrow {OA'}|=|\overrightarrow {OP}|=$radius of the circle 'r'
and $\large\overrightarrow {OA}=-\overrightarrow {OA'}$
To prove that angle in a semicicle is right angle,
We have to prove that angle between AP and A'P is 90$^\circ$
We know from triangle OPA, using triangular law of addition.
$\large\overrightarrow {OP}+\overrightarrow {PA}=\overrightarrow {OA}$
$\Rightarrow \large\overrightarrow {PA}=\overrightarrow {OA}-\overrightarrow {OP}$
Similarly from triangle OPA' we get
$\large\overrightarrow {OP}+\overrightarrow {PA'}=\overrightarrow {OA'}$
$\Rightarrow \large\overrightarrow {PA'}=\overrightarrow {OA'}-\overrightarrow {OP}$
But since $\large\overrightarrow {OA}=-\overrightarrow {OA'}$
$\Rightarrow \large\overrightarrow {PA'}=-\overrightarrow {OA}-\overrightarrow {OP}$
$\large\overrightarrow {PA}.\overrightarrow {PA'}=\big(\overrightarrow {OA}-\overrightarrow {OP}\big).-(\overrightarrow {OA}+\overrightarrow {OP}\big)$
We know that $\large(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\overrightarrow a|^2-|\overrightarrow b|^2$
$=\large|\overrightarrow {OA}|^2-|\overrightarrow {OP}|^2=0$
Because we know that $|\overrightarrow {OA}|=|\overrightarrow {OP}|$
If dot product of two vectors =0 then the angle between them is $90^{\circ}$