When talking about the condition for the three diagonals of a hexagon to be concurrent, we will think of Brianchon's theorem. Using software, I discovered a necessary and sufficient trigonometric relation for the three diagonals to be concurrent, similar to Cava's sine theorem. But I have never seen it anywhere.

Consider $ABCDEF$ be a hexagon, then $AD, BE, CF$ are concurrent if only if

$$\frac{sin(α)} {sin(β)} \frac{sin(γ)} {sin(δ)}\frac{sin(ε)}{sin(ζ)}\frac{sin(η)}{sin(θ)}\frac{sin(ι)}{sin(κ)}\frac{sin(λ)}{sin(μ)}=1$$

Question: Have you seen this result anywhere?

When talking about the condition for the three diagonals of a hexagon to be concurrent, we will think of Brianchon's theorem. Using software, I discovered a necessary and sufficient trigonometric relation for the three diagonals to be concurrent, similar to Cava's sine theorem. But I have never seen it anywhere.

Consider $ABCDEF$ be a hexagon, then $AD, BE, CF$ are concurrent if only if

$$\frac{\sin(α)} {\sin(β)} \frac{\sin(γ)} {\sin(δ)}\frac{\sin(ε)}{\sin(ζ)}\frac{\sin(η)}{\sin(θ)}\frac{\sin(ι)}{\sin(κ)}\frac{\sin(λ)}{\sin(μ)}=1$$

Question: Have you seen this result anywhere?