如图\(1\),已知知矩形\(ABCD\)中,点\(E\)是边\(BC\)上的点,\(AE\)与\(BD\)相交于点\(H\),且\(BE= \sqrt {5},AB=2 \sqrt {5},BC=4 \sqrt {5}\),现将\(\triangle ABD\)沿\(BD\)折起,如图\(2\),点\(A\)的位置记为\(A{{'}}\),此时\(A′E= \sqrt {17}\).
\((1)\)求证:\(BD⊥\)面\(A{{'}}HE\);
\((2)\)求三棱锥\(D-A{{'}}EH\)的体积.