BR1F1SZ wrote: In triangle $ABC$, let $A'$, $B'$ and $C'$ be points on the sides $BC$, $CA$ and $AB$, respectively, such that $$\frac{BA'}{A'C}=\frac{CB'}{B'A}=\frac{AC'}{C'B}.$$ The line parallel to $B'C'$ passing through $A'$ intersects line $AC$ at $P$ and line $AB$ at $Q$. Prove that $$\frac{PQ}{B'C'} \geqslant 2.$$ Let $\frac{BA'}{A'C}=k$. Thus, $$\frac{PQ}{B'C'}=k+\frac{1}{k}\geq2.$$