【Question 30】
In the figure, $B, D, E, C$ lie on the same straight line. It is given that $\angle B A D=\angle D A E=12^{\circ}, A C$ and $A D$ are perpendicular to each other, and $B C=A B+A E$. Find the measure of $\angle A B C$.
【Analysis and Solution】
Geometry, angles.

Extend $BA$ to point $F$, such that $AF=AE$;
Let $\angle ABC=x^{\circ}$;
Since $\angle BAD=\angle DAE=12^{\circ}$;
Therefore, $\angle BAE=\angle BAD+\angle DAE=12^{\circ}+12^{\circ}=24^{\circ}$;
Therefore, $\angle AEC=\angle ABC+\angle BAE=x^{\circ}+24^{\circ}=(x+24)^{\circ}$;
Since $AC$ and $AD$ are perpendicular;
Therefore, $\angle CAD=90^{\circ}$, i.e., $\angle DAE+\angle CAE=90^{\circ}$;
Since $B$, $A$, $F$ are collinear;
Therefore, $\angle BAF=180^{\circ}$, i.e., $\angle BAD+\angle DAE+\angle CAE+\angle CAF=180^{\circ}$;
Therefore, $\angle BAD+\angle CAF=\angle DAE+\angle CAE=90^{\circ}$;
Since $\angle BAD=\angle DAE=12^{\circ}$;
Therefore, $\angle CAF=\angle CAE$;
In $\triangle ACE$ and $\triangle ACF$
$$
\left\{\begin{aligned}
AE & =AF \\
\angle CAE & =\angle CAF; \\
AC & =AC
\end{aligned}\right.
$$

Therefore, $\triangle ACE \cong \triangle ACF$;
Therefore, $\angle AEC=\angle AFC=(x+24)^{\circ}$;
Since $AF=AE$;
Therefore, $AB+AF=AB+AE$, i.e., $BF=AB+AE$;
Since $BC=AB+AE$;
Therefore, $BF=BC$;
Therefore, $\angle BFC=\angle BCF=(x+24)^{\circ}$;
Since $\angle ABC+\angle BFC+\angle BCF=180^{\circ}$;
Therefore, $x^{\circ}+(x+24)^{\circ}+(x+24)^{\circ}=180^{\circ}$;
Solving for $x$ gives $x=44$;
Therefore, $\angle ABC=44^{\circ}$.