Given the right parallelepiped $ABCDA'B'C'D'\ :\ A(0,0,0),B(b,0,0),C(b,d,0),D(0,d,0),A'(0,0,a),B'(b,0,a),C'(b,d,a),D'(0,d,a)$,
$ABCD$ is a rectangle.
Points $E(0,\frac{a^{2}d}{a^{2}+d^{2}},\frac{ad^{2}}{a^{2}+d^{2}})$ and $F(\frac{a^{2}b}{a^{2}+b^{2}+d^{2}},\frac{a^{2}d}{a^{2}+b^{2}+d^{2}},\frac{a(b^{2}+d^{2})}{a^{2}+b^{2}+d^{2}})$.
Equation of the plane $(AEF)\ :\ bx+dy-az=0$.
Points $P(\frac{b^{3}}{b^{2}+d^{2}},\frac{b^{2}d}{b^{2}+d^{2}},a)$ and $Q(\frac{b^{3}}{a^{2}+b^{2}+d^{2}},\frac{b^{2}d}{a^{2}+b^{2}+d^{2}},\frac{a(a^{2}+d^{2})}{a^{2}+b^{2}+d^{2}})$.
Equation of the plane $(B'PQ)\ :\ bx+dy-az=b^{2}-a^{2}$.
$\triangle AEF \sim \triangle B'PQ\ :\ \frac{AE}{B'P}=\frac{AF}{B'Q}=\frac{EF}{PQ}=\frac{a\sqrt{b^{2}+d^{2}}}{b\sqrt{a^{2}+d^{2}}}$.