A few days ago I posted a question and timon92 gave a beautiful solution. Using timon92's method, one can easily show that $ER\cdot BS\cdot CD=DE\cdot RS\cdot BC$ (as with the notations in the figure below). I applied this fact five times and derived the following result.
Let $ABCDE$ be a circumscribed pentagon. $BD$ and $CE$ meet at $P$. $AD$ and $CE$ meet at $Q$. $AD$ and $BE$ meet at $R$. $AC$ and $BE$ meet at $S$. $AC$ and $BD$ meet at $T$. Let $p_1$ be the product of the lengths of the five sides of the pentagon $ABCDE$, $p_2$ be the product of the lengths of the five sides of the pentagon $PQRST$ and $p_3$ be the product of the lengths of the tens sides of the star-shaped decagon $ASBTCPDQER$. Then we have $p_1p_2=p_3$.
The product of the lengths of the red line segments is equal to the product of the lengths of the blue line segments.
My questions:
(1) Are there any other proofs of this result?
(2) Are there any similar results for other star polygons?
Best Answer
Let $p$ be a central projection with a center $A$ from the line $AB$ to the circle.
We see that $p(E)=E$, $p(R)=D$, $p(S)=C$ and $p(B)=B$.
We know that a сross-ratio of $E,$ $S$, $R$ and $B$ on the line $EB$ is equal to the cross-ratio of $E$, $C$, $D$ and $B$ on the circle.
Id est, $$\frac{ER}{EB}:\frac{SR}{SB}=\frac{ED}{EB}:\frac{CD}{CB}$$ or $$\frac{ER\cdot SB}{SR}=\frac{ED\cdot BC}{CD}$$ and we are done.