9. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $A B=A A_{1}$, determine the point $E$ on $B B_{1}$ such that the dihedral angle between the planes $A_{1} E C$ and $A_{1} B_{1} C_{1}$ is $45^{\circ}$.
9. From the given, $\angle E A_{1} B_{1}, \angle C A_{1} C_{1}$ are the angles formed by $A_{1} C, A_{1} E$ with the base plane $A_{1} B_{1} C_{1}$, and $\angle B_{1} A_{1} C_{1}$ is the projection of $\angle E A_{1} C$ on the plane $A_{1} B_{1} C_{1}$. Let $\angle E A_{1} B_{1}=\theta_{1}, \angle C A_{1} C_{1}=\theta_{2}, \angle B_{1} A_{1} C_{1}=\alpha$, and $E B_{1}: B B_{1}=k$, then $\tan \theta_{1}=\frac{E B_{1}}{A_{1} B_{1}}=\frac{E B_{1}}{B B_{1}}=k, \tan \theta_{2}=1$, clearly $a=60^{\circ}$. By Corollary 2 of Property 9, we have $\tan ^{2} 45^{\circ}=\left(k^{2}+1-k\right) /\left(\frac{\sqrt{3}}{2}\right)^{2}$, i.e., $4 k^{2}-4 k+1=0$, which simplifies to $k=\frac{1}{2}$, thus $B E=E B_{1}$. Therefore, when point $E$ is the midpoint of $B B_{1}$, the dihedral angle between the plane $A_{1} E C$ and the plane $A_{1} B_{1} C_{1}$ is $45^{\circ}$.