9. As shown in the figure, in triangle $\mathrm{ABC}$, point $\mathrm{D}$ is on $\mathrm{BC}$, and $\angle \mathrm{ABC}=\angle \mathrm{ACB}$, $\angle \mathrm{ADC}=\angle \mathrm{DAC}$, $\angle \mathrm{DAB}=21^{\circ}$. Find the measure of $\angle \mathrm{ABC}$; and answer: Which triangles in the figure are acute triangles.
9. Solution: $\because \angle \mathrm{DAC}+\angle \mathrm{ADC}+\angle \mathrm{C}=$, and $\angle \mathrm{DAC}=\angle \mathrm{ADC}=\angle \mathrm{B}+21, \angle \mathrm{B}=\angle \mathrm{C}$,
$$
\begin{array}{l}
\therefore 3 \times \angle \mathrm{B}+21^{\circ}=180^{\circ}, \therefore \angle \mathrm{B}=46^{\circ} \\
\angle \mathrm{DAC}=46^{\circ}+21^{\circ}=67^{\circ}, \angle \mathrm{BAC}=67^{\circ}+21^{\circ}=88^{\circ}
\end{array}
$$
$\therefore \triangle \mathrm{ABC}$ and $\triangle \mathrm{ADC}$ are both acute triangles.