Hint for show that $\sin AOB \sin COD + \sin AOC \sin DOB + \sin AOD \sin BOC = 0$

Question

I need a hint for proof of:

If $OA, OB, OC, OD$ any four lines meeting in a point $O$, show that $$\sin AOB \sin COD + \sin AOC \sin DOB + \sin AOD \sin BOC = 0$$

I'm thinking of applying several times the theorem of generalized bisector

Answer 1

Let $\widehat{AOB}=\alpha$, $\widehat{BOC}=\beta$, $\widehat{COD}=\gamma$, $\widehat{DOA}=\delta$.

Then $\alpha+\beta+\gamma+\delta=2\pi$, so, by eliminating $\delta$ and using Briggs' formulas:

$$\begin{eqnarray*} &&2\cdot[ \sin(\alpha)\sin(\gamma)-\sin(\alpha+\beta)\sin(\beta+\gamma)-\sin(\delta)\sin(\beta)]\\&=&2\cdot[\sin(\alpha)\sin(\gamma)-\sin(\alpha+\beta)\sin(\beta+\gamma)+\sin(\alpha+\beta+\gamma)\sin(\beta)]\\&=&\cos(\alpha-\gamma)-\cos(\alpha+\gamma)-\cos(\alpha-\gamma)+\cos(\alpha+2\beta+\gamma)+\cos(\alpha+\gamma)-\cos(\alpha+2\beta+\gamma)\\&=&\color{red}{0}.\end{eqnarray*} $$