The problem is: If on the sides $AB, BC, CD, DA$ of a $\square ABCD$ points $A', B', C', D'$ are taken such that $\overline{AA'}\cdot \overline{BB'}\cdot \overline{CC'}\cdot \overline{DD'}= \overline{A'B} \cdot \overline{B'C} \cdot \overline{C'D} \cdot\overline{D'A}$, show that $A'B'$ and $C'D'$ intersect on $AC$, and $A'D'$ and $B'C'$ intersect on $BD$.
This is a geometry class, we learned Menelaus theorem, Ceva's theorem, Desargue's Theorem(copolar, coaxial) and Pascal's Mystic Hexagram. I draw the picture.
Your given
$$ AA'\times BB'\times CC'\times DD'= A'B \times B'C \times C'D \times D'A $$
can be rewritten as $$\frac{A^{}A'}{A'B^{}}\times\frac{B^{}B'}{B'C^{}}\times\frac{C^{}C'}{C'D^{}}\times\frac{D^{}D'}{D'A^{}}=1.\qquad \qquad (1)$$
Let $A'D'$ intersect $DB$ at $F$ and $B'C'$ intersect $DB$ at $E.$
By Menelaus's theorem we have
$$ \frac{A^{}A'}{A'B^{}}\times\frac{BF}{FD}\times\frac{D^{}D'}{D'A^{}}=-1 \Rightarrow \frac{A^{}A'}{A'B^{}}\times\frac{D^{}D'}{D'A^{}}=\frac{FD}{FB}. \qquad \qquad (2) $$
and
$$ \frac{C^{}C'}{C'D^{}}\times\frac{DE}{EB}\times\frac{B^{}B'}{B'C^{}}=-1 \Rightarrow \frac{C^{}C'}{C'D^{}}\times\frac{B^{}B'}{B'C^{}}=\frac{EB}{ED}. \qquad \qquad (3) $$
Combining $(1),(2)$, and $(3)$ we get
$$ \frac{FD}{FB} \times \frac{EB}{ED} = 1. $$
Using $FB=FD+DB$ and $EB=ED+DB$ and a little manipulation we find that $FD=ED$, hence $F=E$ and $A'D'\cap DB \cap B'C'= F.$
By a similar argument $C'D'$ and $A'B'$ meet on $AC.$