7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $A_{1} C_{1}$ intersects the edges $B_{1} C_{1}$ and $A_{1} B_{1}$ at points $N_{1}$ and $K_{1}$, respectively, which are different from the vertices of the prism. Segments $A N_{1}$ and $C K_{1}$ intersect at point $F$, and at the same time, $A N_{1}=7, C F=4$.

a) Find the angle $F N_{1} B_{1}$.

b) Find the ratio $C_{1} N_{1}: N_{1} B_{1}$.

c) Suppose it is additionally known that $B C=8$. Find the volume of the prism.
Answer: a) $90^{\circ}$, b) $1: 3$, c) $V=56 \sqrt{3}$.

Solution. a) Points $K_{1}$ and $N_{1}$ lie on a circle with diameter $A_{1} C_{1}$; therefore, $\angle A_{1} K_{1} C_{1}=90^{\circ}, \angle A_{1} N_{1} C_{1}=90^{\circ}$ (i.e., $A_{1} N_{1}$ and $C_{1} K_{1}$ are altitudes of triangle $A_{1} B_{1} C_{1}$). The line $A_{1} N_{1}$ is the projection of the line $A N_{1}$ onto the base plane, and since $A_{1} N_{1} \perp B_{1} C_{1}$, by the theorem of three perpendiculars, $A N_{1} \perp B_{1} C_{1}$, i.e., $\angle F N_{1} B_{1}=90^{\circ}$.

b) Since the lines $A N_{1}$ and $C K_{1}$ intersect, all four points $N_{1}, K_{1}, A$, and $C$ lie in the same plane (let's call it $\alpha$). Therefore, the lines $A C$ and $K_{1} N_{1}$ lie in the same plane $\alpha$, and since they do not intersect (as they lie in the parallel bases of the prism), $A C \| K_{1} N_{1}$. Thus, $K_{1} N_{1} \| A_{1} C_{1}$. The trapezoid $A_{1} K_{1} N_{1} C_{1}$ is inscribed in a circle, so it is isosceles, and the angles at its base $A_{1} C_{1}$ are equal, making triangle $A_{1} B_{1} C_{1}$ isosceles ($A_{1} B_{1}=B_{1} C_{1}$).

Triangles $K_{1} N_{1} F$ and $A C F$ are similar by two angles. From the equality of triangles $C C_{1} K_{1}$ and $A A_{1} N_{1}$, it follows that $A N_{1}=C K_{1}$, so both triangles $K_{1} N_{1} F$ and $A C F$ are isosceles with bases $K_{1} N_{1}$ and $A C$ respectively. Therefore, $\quad K_{1} N_{1}: A_{1} C_{1}=K_{1} N_{1}: A C=K_{1} F: C F=\left(K_{1} C-C F\right): C F=\left(A N_{1}-C F\right): C F=3: 4, \quad$ from which $\frac{C_{1} N_{1}}{N_{1} B_{1}}=\frac{C_{1} B_{1}-N_{1} B_{1}}{N_{1} B_{1}}=\frac{C_{1} B_{1}}{N_{1} B_{1}}-1=\frac{A_{1} C_{1}}{N_{1} K_{1}}-1=\frac{4}{3}-1=\frac{1}{3}$.

c) If $B C=8$, then $C_{1} N_{1}=2, B_{1} N_{1}=6, A_{1} B_{1}=8 ; A_{1} N_{1}=\sqrt{A_{1} B_{1}^{2}-B_{1} N_{1}^{2}}=2 \sqrt{7} ; A A_{1}=\sqrt{A N_{1}^{2}-A_{1} N_{1}^{2}}=\sqrt{21}$. Therefore, the area of the base of the prism is $S_{A_{1} B_{1} C_{1}}=\frac{1}{2} \cdot 8 \cdot 2 \sqrt{7}$, and the volume of the prism is $V=S_{A_{1} B_{1} C_{1}} \cdot A A_{1}=56 \sqrt{3}$.