Given a rectangular parallelepiped $ABCDA_1B_1C_1D_1$. Let the points $E$ and $F$ be the feet of the perpendiculars drawn from point $A$ on the lines $A_1D$ and $A_1C$, respectively, and the points $P$ and $Q$ be the feet of the perpendiculars drawn from point $B_1$ on the lines $A_1C_1$ and $A_1C$, respectively. Prove that $\angle EFA = \angle PQB_1$