1. As shown in Figure 1, in $\triangle A B C$, $\angle A>90^{\circ}, D$ is a point on side $B C$ such that $A B$ is tangent to the circumcircle of $\triangle A C D$ at point $A$. The projection of point $B$ onto line $A D$ is $E$. If $C A=C D, A E=C E$, find the measure of $\angle B C A$.
1. Let $\angle B C A=2 \alpha$. Then
$\angle C A D=\angle C D A=90^{\circ}-\alpha, \angle B A D=2 \alpha$.
Since $A E=C E$, we have $\angle A E C=2 \alpha$.
Thus, $\triangle A E C \backsim \triangle A C D \Rightarrow \frac{A E}{A C}=\frac{A C}{A D}$.
Let $B C=a, C A=b, A B=c$.
Since $B E \perp A E$, we have
$A E=A B \cos 2 \alpha=c \cos 2 \alpha$.
In $\triangle A C D$, by the Law of Sines,
$$
\begin{array}{l}
\frac{A D}{\sin 2 \alpha}=\frac{A C}{\sin \left(90^{\circ}-\alpha\right)} \Rightarrow A D=2 b \sin \alpha \\
\Rightarrow b^{2}=A C^{2}=A E \cdot A D=c \cos 2 \alpha \cdot 2 b \sin \alpha .
\end{array}
$$

Let the circumradius of $\triangle A B C$ be $R$.
By the Law of Sines, we get $b=2 R \sin B, c=2 R \sin C$.
$$
\begin{array}{l}
\text { Also, } \angle A B C=180^{\circ}-\angle B A C-\angle A C B \\
=180^{\circ}-\left(90^{\circ}-\alpha+2 \alpha\right)-2 \alpha \\
=90^{\circ}-3 \alpha,
\end{array}
$$

Thus, $b=2 R \cos 3 \alpha, c=2 R \sin 2 \alpha$.
Hence, $\cos 3 \alpha=2 \sin 2 \alpha \cdot \cos 2 \alpha \cdot \sin \alpha$
$$
\begin{array}{l}
=\sin 4 \alpha \cdot \sin \alpha . \\
\text { By } \cos 3 \alpha=\cos (4 \alpha-\alpha) \\
=\cos 4 \alpha \cdot \cos \alpha+\sin 4 \alpha \cdot \sin \alpha \\
\Rightarrow \cos 4 \alpha \cdot \cos \alpha=0 .
\end{array}
$$

This indicates that $\alpha=90^{\circ}$ or $4 \alpha=90^{\circ}$.
If $\alpha=90^{\circ}$, then $\angle B C A=180^{\circ}$, which is a contradiction.
Thus, $4 \alpha=90^{\circ}$.
Therefore, $\angle B C A=2 \alpha=45^{\circ}$.