<s>[INST] 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:
$$y-c\sin \beta=\dfrac {c\sin \alpha - c\sin \beta}{c\cos \alpha- c\cos \beta} (x-c\cos \beta)$$
$$y-c\sin \beta =\dfrac {\sin \alpha - \sin \beta}{\cos \alpha - \cos \beta} (x-c\cos \beta)$$
$$x(\sin \alpha - \sin \beta)-y(\cos \alpha - \cos \beta)=c \sin \alpha. \cos \beta - c \cos \alpha. \sin \beta$$
$$x(\sin \alpha - \sin \beta) - y (\cos \alpha - \cos \beta)= c\sin (\alpha - \beta)$$
 [/INST] Hint:
Using Prosthaphaeresis Formulas
The gradient of $$A(c\cos\alpha,c\sin\alpha);B(c\cos\beta,c\sin\beta)$$
is $$\dfrac{\sin\alpha-\sin\beta}{\cos\alpha-\cos\beta}=-\cot\dfrac{\alpha+\beta}2$$  assuming  $\sin\dfrac{\alpha-\beta}2\ne0$ as for $\sin\dfrac{\alpha-\beta}2=0,\alpha\equiv\beta\pmod{2\pi}\implies A,B$ coincide.
The midpoint$(M)$ of $$(c\cos\alpha,c\sin\alpha);(c\cos\beta,c\sin\beta)$$
is  $$\left(\dfrac{c(\cos\alpha+\cos\beta)}2,\dfrac{c(\sin\alpha+\sin\beta)}2\right)$$
So, the gradient of $O(0,0);M$ will be $$\dfrac{\dfrac{c(\sin\alpha+\sin\beta)}2-0}{\dfrac{c(\cos\alpha+\cos\beta)}2-0}=\tan\dfrac{\alpha+\beta}2$$
assuming  $\cos\dfrac{\alpha-\beta}2\ne0$ as for $\cos\dfrac{\alpha-\beta}2=0,\alpha\equiv\pi+\beta\pmod{2\pi}\implies A,B$ becomes extremities of a diameter .
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