Let $O$ be any point on the circumference of a circle circumscribing a regular polygon $A_1,A_2,A_3.., A_{2n+1}$ such that $O$ lies on the arc $ A_1A_{2n+1} $. Show that $OA_1+OA_3+...OA_{2n+1}=OA_2+OA_4...+OA_{2n} $.
I tried by find the length of the sides from the angles $OPA_1 , OPA_2 , OPA_3 ... , OPA_{2n+1}$ Where i considered $P$ to be the circle with ,$OPA_1 = \theta$ and the radius of this circle to be $r$ .
Solving further i got struck and got a different answer. Can you pls help me solve this question in this method, If you have some other approach to this question then please share it.
Thank You
The key formula that we need can be found HERE:
$$\sum_{k=0}^n\sin(\varphi + k\alpha)=\frac{\sin\frac{(n+1)\alpha}{2}\sin(\varphi+\frac{n\alpha}2)}{\sin\frac\alpha2}$$
If you apply the following identity:
$$\sin\alpha\sin\beta=\frac12(cos(\alpha-\beta)-cos(\alpha+\beta))$$
...you get:
$$\sum_{k=0}^n\sin(\varphi + k\alpha)=\frac{\cos(\frac\alpha2-\varphi)-\cos(\frac{2n+1}{2}\alpha+\varphi)}{2\sin\frac\alpha2}\tag{1}$$
Now, denote the center of circumscribed circle with $C$. Introduce:
$$\angle OCA_{2n+1}=-2\varphi$$
$$\angle A_kCA_{2n+1}=k\alpha$$
$$\alpha=\frac{2\pi}{2n+1}\tag{2}$$
You can easily show that:
$$OA_k=2R\sin\frac{\angle A_kCA_{2n+1}-\angle OCA_{2n+1}}2=2R\sin(k\frac\alpha2+\varphi)$$
Now calculate two sums:
$$OA_1+OA_3+...OA_{2n+1}=2R\sin(\frac\alpha2+\varphi)+2R\sin(\frac{3\alpha}2+\varphi)+...+2R\sin(\frac{(2n+1)\alpha}2+\varphi)\tag{3}$$
$$OA_2+OA_4+...OA_{2n}=2R\sin(\alpha+\varphi)+2R\sin(2\alpha+\varphi)+...+2R\sin(n\alpha+\varphi)\tag{4}$$
Now apply formula (1) on (3) and (4) and simplify by taking into account (2). It should not be too complicated to show that (3) and (4) are actually equal.