Example 5 (2003 Belarusian Mathematical Olympiad) Given a cyclic quadrilateral $A B C D$ satisfying $A B=B C=A D$ $+C D, \angle B A D=\alpha, A C=d$. Find the area of $\triangle A B C$.
Analysis of solving by parameter method.
Solution Extend $A D$ to $M$, such that $D M=D C$,
Let $A D=x, C D=y, A B=z$.
Since $\angle C D M=\angle A B C$
$$
\frac{C D}{A B}=\frac{D M}{B C}
$$

Therefore, $\triangle D C M \sim \triangle B A C$
Let $k=\frac{C D}{A B}=\frac{y}{z}$, then
$$
\begin{array}{l}
C M=k \cdot A C=k d \\
S_{\triangle C D M}=k^{2} S_{\triangle A B C}
\end{array}
$$

Also, $S_{\triangle C D M}=\frac{y}{x+y} S_{\triangle A C M}=\frac{y}{z} S_{\triangle A C M}=k S_{\triangle A C M}$
Thus, $S_{\triangle A B C}=\frac{1}{k^{2}} S_{\triangle C D M}=\frac{1}{k^{2}} \cdot k S_{\triangle A C M}=\frac{1}{k} \cdot \frac{1}{2} A C \cdot C M \sin \angle A C M$
Since $\angle D C M=\angle B C A$
Therefore, $\angle A C M=\angle B C D=180^{\circ}-\alpha$
Thus, $S_{\triangle A B C}=\frac{1}{k} \cdot \frac{1}{2} d \cdot k d \sin \left(180^{\circ}-\alpha\right)=\frac{1}{2} d^{2} \sin \alpha$