\(BB'//A'A\Rightarrow d\left(BB';A'H\right)=d\left(BB';\left(A'AH\right)\right)=d\left(B;\left(A'AH\right)\right)\)
\(\left\{{}\begin{matrix}A'H\perp BC\\BC\perp AH\end{matrix}\right.\) \(\Rightarrow BC\perp\left(A'AH\right)\Rightarrow BH=d\left(B;\left(A'AH\right)\right)\)
\(BH=\frac{1}{2}BC=a\)