parmenides51 wrote: Let $ABCDEF$ be a convex hexagon, $P, Q, R$ be the intersection points of $AB$ and $EF$, $EF$ and $CD$, $CD$ and $AB$. $S, T,UV$ are the intersection points of $BC$ and $DE$, $DE$ and $FA$, $FA$ and $BC$, respectively. Prove that if $$\frac{AB}{PR}=\frac{CD}{RQ}=\frac{EF}{QP},$$then $$\frac{BC}{US}=\frac{DE}{ST}=\frac{FA}{TU}.$$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $L_1$ be parallel to $RQ$ passing through $B$ and $L_2$ be parallel to $PQ$ passing through $A$ Let $X\equiv L_1 \cap L_2$ $$\Rightarrow \triangle ABX \equiv \triangle PRQ$$$$\Rightarrow BX=CD \text{ and } AX=FE$$$$\Rightarrow AFEX \text{ and } BCDX \text{ are parallelograms}$$$$\Rightarrow BC=XD \text{ and }AF=XE , US//XD \text{ and } UT//XE$$$$\Rightarrow \triangle XED \equiv \triangle UTS$$$$\Rightarrow \frac{XD}{US}=\frac{DE}{ST}=\frac{XE}{TU}$$$$\Rightarrow \frac{BC}{US}=\frac{DE}{ST}=\frac{FA}{TU}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$