Question: Suppose that the sides $AB, BC, CD$ and $DA$ of a quadrilateral $ABCD$ are cut by a line at the points $A' B' C' D'$ respectively, show that $AA'/A'B \times BB'/B'C \times CC'/C'D \times DD'/D'A = +1$

Question (harder geometry):
Suppose that the sides $AB, BC, CD$ and $DA$ of a quadrilateral $ABCD$ are cut by a line at the points $A' B' C' D'$ respectively, show that 
$$AA'/A'B \times BB'/B'C \times CC'/C'D \times DD'/D'A = +1$$

SOLUTION:  ================


Referring to my diagram, see attached.

I simply used the sine rule 4 times on triangles $A'BB', CB'C', C'DD', AA'D$ ;

let the angles be $\alpha, \beta, \gamma, \delta$.

$A'B/\sin(\pi-\alpha-\beta) = BB'/\sin \alpha$ (1)
$C'D/\sin \delta = DD'/sin (\pi-\gamma-\delta)$ (2)
$B'C/\sin(\pi-\gamma-\delta) = CC'/\sin(\pi-\alpha-\beta)$ (3)
$AA/\sin(\pi-\delta) = AD'/\sin (\pi-\alpha)$ (4)

so
$BB'/A'B = \sin \alpha/ \sin(\pi-\alpha-\beta) $
$DD'/C'D = \sin (\pi-\gamma-\delta)/\sin \delta $
$CC'/B'C = \sin(\pi-\alpha-\beta)/\sin(\pi-\gamma-\delta)$
$AA'/AD' = \sin(\pi-\delta)/\sin(\pi-\alpha) = \sin \delta/sin \alpha $
[since $\sin(A-B)=\sin A \cos B - \cos B \sin A, and \cos \pi = -1$]

Then we have

$AA'/A'B \times BB'/B'C \times CC'/C'D \times DD'/D'A $
$= BB'/A'B\times CC'/B'C\times DD'/C'D\times AA'/D'A $
$={\sin\alpha \over \sin(\pi-\alpha-\beta)}\times {\sin(\pi-\alpha-\beta)\over \sin(\pi-\gamma-\delta)}\times {\sin (\pi-g-\delta)\over \sin \delta}\times {\sin \delta\over \sin \alpha}$

$= 1$ [since everything cancels to 1.]
as required.//