Local Mathematical Competitions

The given relation writes as $\frac{BF}{F{C}^{\prime }}=\frac{B^{\prime }E}{EC}$. Let $M$ be a point on the side $BC$ such that $FM\parallel C{C}^{\prime }$. Then $\frac{BM}{MC}=\frac{BF}{F{C}^{\prime }}=\frac{B^{\prime }E}{EC}$, implying $ME\parallel B{B}^{\prime }$. Hence the angles $\angle AEM$, $\angle AFM$ and $\angle A{A}^{\prime }M$ are right angles, therefore $E$, $F$, $A^{\prime }$ all lie on the circle of diameter $AM$. The conclusion follows.