It's well known that: $A,H',B,C$ are concyclic. Let $Q=BH \cap AC$ $\implies \angle ACB=\angle BCF=\angle AHQ=\angle FHQ=\frac{\angle AHF}{2}=\angle AEF$ $\implies BEHFC$ is cyclic. By symmetry $B,E',H',F',C$ es cyclic. Then: $A,E',F',H$ is concyclic.$\blacksquare$

$F$ is the symmetric point of $A$ over the foot from $B$ to $AC$, so $\measuredangle BFA=\measuredangle FAB$. so $\measuredangle BFC=\measuredangle BFA=\measuredangle FAB=\measuredangle CAB=\measuredangle BHC$. Now we have $BHFC$ is cyclic, so similarly $BEFC$ is cyclic. As $(BHC)$ is the reflection of $(ABC)$ over $BC$, we know that $E', F', H'$ lies on the circumcircle, we're done.