Let $ABCDEF$ be a convex hexagon such that $\angle B+\angle D+\angle F=360^{\circ }$and
$\frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1$ .
Prove that $\frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1$.
I know it is an old but nice question it belongs to IMO shortlist in 1998.you can find out some proof in below link but unfortunately there is no synthetic solution in the link
https://artofproblemsolving.com/community/c6h1117p3488
I think we should consider this fact that the circumcircles of triangles $ABC, CDE, EFA$ have a common point $P$.
Because, if $P$ is the intersection of circumcircles $CDE, EFA$ , then:
$\angle EPC=180-\angle D$
$\angle APE=180-\angle F$
then: $\angle APC=360-(\angle D+\angle F)=\angle B$
and it means that $P$ is on circumcircle of $ABC$ ,too.
Please post synthetic answers. Thanks!
OFFICIAL SOLUTION:
Take P so that AFE and PDE are similar. [So take ∠PDE = ∠AFE and ∠PED = ∠AEF. Note that you need to take the right point - P' fails below.] So EF/ED = EA/EP. Also ∠DEF = ∠DEA + ∠AEF = ∠DEA + ∠PED = ∠PEA (this is where we need P on the opposite side of DE to A). Hence triangles DEF and PEA are similar. So FD/EF = AP/EA (*).
Triangles AFE, PDE similar also gives FA/EF = DP/DE. So AB/BC = (DE/CD) (FA/EF) (given) = (DE/CD) (DP/DE) = PD/DC. But ∠CDE + ∠EDP + ∠PDC = 360o, or ∠D + ∠F + ∠PDC = 360o. We are given that ∠B + ∠D + ∠F = 360o, so ∠PDC = ∠ABC. Hence triangles ABC and PDC are similar. So CB/CD = CA/CP. ∠BCD = ∠ACD + ∠BCA = ∠ACD + ∠DCP = ∠ACP. So triangles BCD and ACP are similar. Hence BC/BD = AC/AP. Multiplying by (*) gives (BC/BD) (FD/EF) = AC/EA, which is the required equality