Prove that the perpendicular from the origin upon the straight line Prove that the perpendicular drawn from the origin upon the straight line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ bisects the distance between them.

My Attempt:
Equation of the line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ is:
$$y-c\sin \beta=\dfrac {c\sin \alpha - c\sin \beta}{c\cos \alpha- c\cos \beta} (x-c\cos \beta)$$
$$y-c\sin \beta =\dfrac {\sin \alpha - \sin \beta}{\cos \alpha - \cos \beta} (x-c\cos \beta)$$
$$x(\sin \alpha - \sin \beta)-y(\cos \alpha - \cos \beta)=c \sin \alpha. \cos \beta - c \cos \alpha. \sin \beta$$
$$x(\sin \alpha - \sin \beta) - y (\cos \alpha - \cos \beta)= c\sin (\alpha - \beta)$$
The two points $A$ and $B$ lie on the circle of radius $c$ centred on the origin $O$, having arguments $\alpha$ and $\beta$ respectively. Now consider $\triangle AOB$; because $AO=OB$ the triangle is isosceles and the perpendicular to $AB$ through $O$ bisects $AB$. This completes the proof.
Prove that the perpendicular bisector  from origin to the line joining the points $(c\cos\alpha,c\sin\alpha)$ and $(c\cos\beta,c\sin\beta)$ passes through the center as well.