Let $ABCD$ be a convex quadrilateral and $X$ is an interior point.
Also let $AX\cap BD=\{E\}$, $BX\cap AC=\{F\}$, $CX\cap BD=\{G\}$ and $DX\cap AC=\{H\}$.
Prove that:
$$AF\cdot BG\cdot CH\cdot DE=AH\cdot BE\cdot CF\cdot DG$$
I proved it by the following ugly reasoning.
Let $AC\cap BD=\{O\}$, $AH:HO:OC=x:y:z$ and $BO:OG:GD=a:b:c$.
Hence, by Thales we can get that:
$$\frac{AF}{AH}\cdot\frac{CH}{CF}=\frac{(x+y)(y+z)(a+b)(b+c)}{acy(x+y+z)+(a+b)(b+c)xz}$$ and $$\frac{BG}{BE}\cdot\frac{GE}{DG}=\frac{xzb(a+b+c)+(x+y)(y+z)ac}{(x+y)(y+z)(a+b)(b+c)}$$
and we are done.
Is there something easy here?
Thank you!