Example 1 On the Cartesian plane, given the parabola $y=1-x^{2}$ and the line $y=x+a(-1<a<1)$ intersect at points $A$ and $B$, and point $C(1,0)$. Question: For what value of $a$ is the area of $\triangle ABC$ maximized? Find the maximum area of $\triangle ABC$. ${ }^{[1]}$
$(2010$, Shanghai $\mathrm{TI}$ Cup High School Grade 2 Mathematics Competition)
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y=1-x^{2}, \\
y=x+a
\end{array}\right. \\
\Rightarrow x^{2}+x+a-1=0 \\
\Rightarrow x_{1}+x_{2}=-1, x_{1} x_{2}=a-1 \\
\Rightarrow|A B|^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=2\left(x_{1}-x_{2}\right)^{2} \\
\quad=2\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]=2(5-4 a) .
\end{array}
$$

The distance from point $C$ to the line $A B$ is $d=\frac{|1+a|}{\sqrt{2}}$, then
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2}|A B| d \\
=\frac{1}{2} \sqrt{(5-4 a)(1+a)^{2}} \\
=\frac{1}{2} \sqrt{2\left(\frac{5}{2}-2 a\right)(1+a)(1+a)} \\
\leqslant \frac{1}{2} \sqrt{2\left[\frac{\left(\frac{5}{2}-2 a\right)+(1+a)+(1+a)}{3}\right]^{3}} \\
=\frac{3 \sqrt{3}}{4} .
\end{array}
$$

When and only when $\frac{5}{2}-2 a=1+a$, i.e., $a=\frac{1}{2}$, $S_{\triangle A B C}$ reaches its maximum value $\frac{3 \sqrt{3}}{4}$.