10. (20 points) Let the left vertex of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $A$, and the right focus be $F(c, 0)$, and $2b$, $a$, $c$ form a geometric sequence. A line passing through point $F$ intersects the ellipse at points $M$ and $N$, and the lines $AM$ and $AN$ intersect the right directrix at points $P$ and $Q$, respectively. Find the size of $\angle P F Q$.
10. From $a^{2}=2 b c \Rightarrow a^{4}=4\left(a^{2}-c^{2}\right) c^{2}$.

Solving gives $e=\frac{\sqrt{2}}{2}, a^{2}=2 b^{2}=2 c^{2}$.
Then the equation of the ellipse can be written as $x^{2}+2 y^{2}-2 c^{2}=0$.
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$,
$l_{M N}: y=k(x-c)$.
Combining with the ellipse equation, eliminating $y$ gives
$$
\left(2 k^{2}+1\right) x^{2}-4 k^{2} c x+2 c^{2}\left(k^{2}-1\right)=0 \text {. }
$$

Then $x_{1}+x_{2}=\frac{4 k^{2} c}{2 k^{2}+1}, x_{1} x_{2}=\frac{2 c^{2}\left(k^{2}-1\right)}{2 k^{2}+1}$.
From $l_{M A}: y=\frac{y_{1}}{x_{1}+a}(x+a)$ we get
$$
P\left(2 c, \frac{y_{1}}{x_{1}+a}(2 c+a)\right) \text {, }
$$

From $l_{N A}: y=\frac{y_{2}}{x_{2}+a}(x+a)$ we get
$$
\begin{array}{l}
Q\left(2 c, \frac{y_{2}}{x_{2}+a}(2 c+a)\right) . \\
\text { Hence } \overrightarrow{F P}=\left(c, \frac{y_{1}}{x_{1}+a}(2 c+a)\right), \\
\overrightarrow{F Q}=\left(c, \frac{y_{2}}{x_{2}+a}(2 c+a)\right) .
\end{array}
$$

Then $\overrightarrow{F P} \cdot \overrightarrow{F Q}$
$$
\begin{array}{l}
=c^{2}+\frac{y_{1} y_{2}}{\left(x_{1}+a\right)\left(x_{2}+a\right)}(2 c+a)^{2} \\
=c^{2}+\frac{k^{2}\left(x_{1} x_{2}-c\left(x_{1}+x_{2}\right)+c^{2}\right)}{x_{1} x_{2}+a\left(x_{1}+x_{2}\right)+a^{2}}(2 c+a)^{2} .
\end{array}
$$

And $k^{2}\left(x_{1} x_{2}-c\left(x_{1}+x_{2}\right)+c^{2}\right)$
$$
\begin{array}{l}
=k^{2}\left(\frac{2 c^{2}\left(k^{2}-1\right)}{2 k^{2}+1}-\frac{4 k^{2} c^{2}}{2 k^{2}+1}+c^{2}\right) \\
=\frac{-k^{2} c^{2}}{2 k^{2}+1} \\
\Rightarrow x_{1} x_{2}+a\left(x_{1}+x_{2}\right)+a^{2} \\
=\frac{2 c^{2}\left(k^{2}-1\right)}{2 k^{2}+1}+\frac{4 k^{2} c a}{2 k^{2}+1}+a^{2} \\
=\frac{2 k^{2}(a+c)^{2}}{2 k^{2}+1} .
\end{array}
$$

From $a^{2}=2 c^{2}$, then
$\overrightarrow{F P} \cdot \overrightarrow{F Q}$
$$
\begin{array}{l}
=c^{2}+\frac{k^{2}\left(x_{1} x_{2}-c\left(x_{1}+x_{2}\right)+c^{2}\right)}{x_{1} x_{2}+a\left(x_{1}+x_{2}\right)+a^{2}}(2 c+a)^{2} \\
=c^{2}+\frac{\frac{-k^{2} c^{2}}{2 k^{2}+1}}{\frac{2 k^{2}(a+c)^{2}}{2 k^{2}+1}}(2 c+a)^{2} \\
=c^{2}-c^{2} \frac{(2 c+a)^{2}}{2(a+c)^{2}} \\
=c^{2}-c^{2}=0 .
\end{array}
$$

Thus $\overrightarrow{F P} \cdot \overrightarrow{F Q}=0$
$$
\Rightarrow \angle P F Q=90^{\circ} \text {. }
$$