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:
$$y-c\sin \beta=\dfrac {c\sin \alpha - c\sin \beta}{c\cos \alpha- c\cos \beta} (x-c\cos \beta)$$
$$y-c\sin \beta =\dfrac {\sin \alpha - \sin \beta}{\cos \alpha - \cos \beta} (x-c\cos \beta)$$
$$x(\sin \alpha - \sin \beta)-y(\cos \alpha - \cos \beta)=c \sin \alpha. \cos \beta - c \cos \alpha. \sin \beta$$
$$x(\sin \alpha - \sin \beta) - y (\cos \alpha - \cos \beta)= c\sin (\alpha - \beta)$$
 A: 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.
A: Those points satisfy the following equation,
$$x^2+y^2=c^2$$
Which is a circle with centre (0,0) and radius $c^2$
Now using the property of circle that

perpendicular drawn from the centre of a circle to any cord on the circle bisects the cord.

Hence perpendicular drawn from origin to the line bisects it.
A: 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' = $
$m' × \dfrac{\cos\beta +\cos\alpha}{\cos\beta + cos\alpha}=$
$\dfrac{(\sin\beta -\sin\alpha)(\cos\beta + \cos\alpha)}{\cos^2 \beta - \cos^2\alpha} =$
$\dfrac{(\sin\beta -\sin\alpha)(\cos\beta +\cos\alpha)}{\sin^2\alpha - \sin^2\beta}=$
$- \dfrac{cos\beta +\cos\alpha}{\sin\beta +sin\alpha}=$
$-(1/m)$.
A: 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 .
A: Let $O$ be the origin. 
Let $A$ be equal to $(c \cos \alpha, c \sin \alpha)$.
Let $B$ be equal to $(c \cos \beta, c \sin \beta)$.
Let $v_A$ denotes $\vec{OA}$ and $v_B$ denotes $\vec{OB}$
Let $d$ denotes the $\vec{AB}=v_B-v_A$.
Let $P$ denotes the projection of $O$ on line $AB$. We want to show that $|AP|=|PB|$.
$$|PA| = \frac{|v_A.d|}{\|d\|}=\frac{|v_A.(v_B-v_A)|}{\|d\|}=\frac{|v_A.v_B-c^2|}{\|d\|}$$
$$|PB| = \frac{|v_B.d|}{\|d\|}=\frac{|v_B.(v_B-v_A)|}{\|d\|}=\frac{|c^2-v_A.v_B|}{\|d\|}$$
Hence $|PA|=|PB|$.
A: 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.
