In a triangle $ABC$ the heights $BB'$ and $CC'$ intersect the circumcircle of $ABC$ at $E$ and $F$. Prove that $\displaystyle B'C'=\frac{EF}{2}$.
The circle of diameter $BC$ passes through $B'$ and $C'$ since $BB'C=CC'B=90^{\circ}$. $EFC=\frac{\stackrel{\frown}{EC}}{2}=B'BC=CC'B$ then the lines $B'C'$ and $EF$ are parallel. They are also parallel to the tangent at $A$. How to continue?