Heights and circumcircle

Question

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?

Answer 1

Let $H$ be the orthocenter of $ABC$. It is well know (and easy to prove) that reflection of $H$ across $B'$ and $C'$ is respectively $E$ and $F$. So $B'C'$ is a middle line in triangle $HEF$ parallel to $EF$, so $B'C' =EF/2$.

Answer 2

Let $\angle CAB=\alpha$, $\angle ABC=\beta$, $\angle BCA=\gamma$.

\begin{align} \angle ABE&=\angle FBA=90^\circ-\alpha ,\\ \angle FBE&=180^\circ-2\alpha ,\\ \triangle BB_1C_1:\quad \frac{|B_1C_1|}{\sin\angle C_1BB_1} &=\frac{|BC_1|}{\sin\angle BB_1C_1} ,\\ \frac{|B_1C_1|}{\cos\alpha} &=\frac{|BC|\cos\beta}{\cos\beta} ,\\ |B_1C_1|&=|BC|\cos\alpha ,\\ \triangle BEF:\quad \frac{|EF|}{\sin2\alpha} &= \frac{|BF|}{\cos\beta} = \frac{|BC|\cos\beta/\sin\alpha}{\cos\beta} = \frac{|BC|}{\sin\alpha} ,\\ |EF|& =\frac{|BC|\sin2\alpha}{\sin\alpha} =\frac{2|BC|\sin\alpha\cos\alpha}{\sin\alpha} =2|BC|\cos\alpha ,\\ |EF|&=2|B_1C_1| . \end{align}