Equal products of triangle areas

Question

Can you prove the following claim:

Claim. Given hexagon circumscribed about an ellipse. Let $A_1,A_2,A_3,A_4,A_5,A_6$ be the vertices of the hexagon and let $B$ be the intersection point of its principal diagonals. Denote area of triangle $\triangle A_1A_2B$ by $K_1$, area of triangle $\triangle A_2A_3B$ by $K_2$,area of triangle $\triangle A_3A_4B$ by $K_3$,area of triangle $\triangle A_4A_5B$ by $K_4$,area of triangle $\triangle A_5A_6B$ by $K_5$ and area of triangle $\triangle A_1A_6B$ by $K_6$ .Then, $$K_1 \cdot K_3 \cdot K_5=K_2 \cdot K_4 \cdot K_6$$

The GeoGebra applet that demonstrates this claim can be found here.

Answer 1

I am using the following fact given a triangle $ABC$ its area is equal to $\frac{1}{2}AB.BC.\text{sin}(\hat{ABC})$. Now use this formula for the both hand sides. First of all note that the fact that $A_2A_5$, $A_1A_4$ and $A_3A_6$ intersect at a point is non-trivial and it is called the Brianchon's theorem. Now we have the following:

$$K_1K_3K_5=\frac{1}{8}BA_1.BA_2.BA_3.BA_4.BA_5.BA_6\text{sin}(\hat{A_1BA_2})\text{sin}(\hat{A_4BA_3})\text{sin}(\hat{A_5BA_6})$$ and similarly we have: $$K_2K_4K_6=\frac{1}{8}BA_1.BA_2.BA_3.BA_4.BA_5.BA_6\text{sin}(\hat{A_2BA_3})\text{sin}(\hat{A_4BA_5})\text{sin}(\hat{A_1BA_6})$$ One can see these two quantities are the same since $\hat{A_1BA_2}=\hat{A_4BA_5}$, $\hat{A_4BA_3}=\hat{A_1BA_6}$ and $\hat{A_5BA_6}=\hat{A_2BA_3}$.