$$
\begin{aligned}
& \text { In } \triangle A C D, \cos C=\frac{b}{\left(\frac{a}{2}\right)}=\frac{2 b}{a} \\
& \text { In } \triangle A B C, \cos C=\frac{a^2+b^2-c^2}{2 a b}
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\begin{array}{rlrl}
& & \frac{2 b}{a} =\frac{a^2+b^2-c^2}{2 a b} \\
\Rightarrow & 4 b^2 =a^2+b^2-c^2 \\
& \Rightarrow & 3 b^2 =a^2-c^2 \\
& \Rightarrow & b^2 =\frac{a^2-c^2}{3}
\end{array}
$$


Also,
$$
\text { Also, } \quad \begin{aligned}
\cos A & =\frac{b^2+c^2-a^2}{2 b c} \\
\therefore \quad \cos A \cos C & =\frac{b^2+c^2-a^2}{2 b c} \times \frac{2 b}{a} \\
& =\frac{b^2+c^2-a^2}{a c} \\
& =\frac{\frac{1}{3}\left(a^2-c^2\right)+\left(c^2-a^2\right)}{a c} \\
& =\frac{a^2-c^2+3 c^2-3 a^2}{3 a c} \\
& =\frac{2 c^2-2 a^2}{3 a c}=\frac{2\left(c^2-a^2\right)}{3 a c}
\end{aligned}
$$