<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] Let $O$ be the origin.
Midpoint $M$ of $A$ and $B$:
$x_M =(1/2)c(\cos\alpha +\cos\beta)$;
$y_M =(1/2)c(\sin\alpha + \sin\beta)$;
Line $l_1$  joining points $O$(origin) and $M$ has slope:
$m: = \dfrac{y_M}{x_M}$.
Line $l_2$ joining $A$ and $B$ has slope:
$m':= \dfrac{\sin\beta - sin\alpha}{\cos\beta - \cos\alpha}$.
Remains to be shown:
$m'= -\dfrac{1}{m}$.
$m' = $
$m' × \dfrac{\cos\beta +\cos\alpha}{\cos\beta + cos\alpha}=$
$\dfrac{(\sin\beta -\sin\alpha)(\cos\beta + \cos\alpha)}{\cos^2 \beta - \cos^2\alpha} =$
$\dfrac{(\sin\beta -\sin\alpha)(\cos\beta +\cos\alpha)}{\sin^2\alpha - \sin^2\beta}=$
$- \dfrac{cos\beta +\cos\alpha}{\sin\beta +sin\alpha}=$
$-(1/m)$.
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