A transversal on the sides $AB, BC, CD, DA$ of a quadrilateral $ABCD$ at the points $P, Q, R, S$ respectively. Prove that $\frac{AP}{PB}·\frac{BQ}{QC}·\frac{CR}{RD}·\frac{DS}{SA} = +1$
My first attempt was to prove this by using Menelaus's theorem but couldn't come to a satisfactory result. I would be highly obliged for a perfect solution to this problem. Thank you!
First, we draw a quadrilateral ABCD and draw the transversal that cuts the quadrilateral as instructed in the aforementioned question.
Next, we connect AC.
Let the point of intersection of the transversal and the line AC be T.
On applying Menelaus' theorem in $\Delta ABC$, we get:
$\frac{BQ}{QC}·\frac{TC}{TA}·\frac{AP}{PB}=-1$ (1)
Similarly, on applying Menelaus' theorem in $\Delta ADC$, we get:
$\frac{DS}{SA}·\frac{AT}{TC}·\frac{CR}{RD}=-1$ (2)
Now, Multiplying (1) and (2):
$\frac{BQ}{QC}·\frac{TC}{TA}·\frac{AP}{PB}·\frac{DS}{SA}·\frac{AT}{TC}·\frac{CR}{RD}=1$
Hence, $\frac{BQ}{QC}·\frac{AP}{PB}·\frac{DS}{SA}·\frac{CR}{RD}=+1$
Proved!