Prove that $\frac{AP}{PB} \frac{BQ}{QC} \frac{CR}{RD} \frac{DS}{SA}=+1$.

Question

A transversal cuts the sides $AB,BC,CD,DA$ of a quadrilateral $ABCD$ at $P,Q,R,S$ respectively. Prove that $\frac{AP}{PB} \frac{BQ}{QC} \frac{CR}{RD} \frac{DS}{SA}=+1$.

I can only think of these two equations:$$\frac{BQ}{XQ} \frac{XS}{SA} \frac{AP}{PB} =+1$$ and $$\frac{BP}{PY} \frac{YC}{RC} \frac{CQ}{QB} =+1$$ What next?

Answer 1

Apply Menelaus's theorem to triangles $ABD$ and $CDB$.