20. Suppose that $A B C$ is a triangle and $D$ is a point on side $A B$ with $A D=B D=C D$. If $\angle \mathrm{ACB}=x^{\circ}$, find the value of $x$.
20. Answer: 90

First note that since $\triangle \mathrm{DAC}$ and $\triangle \mathrm{DBC}$ are isosceles triangles, we have $\angle \mathrm{DAC}=\angle \mathrm{DCA}$ and $\angle \mathrm{DBC}=\angle \mathrm{DCB}$. Now by considering the sum of angles in $\triangle \mathrm{ABC}$, we have
$$
\begin{array}{l}
\angle \mathrm{DAC}+\angle \mathrm{DBC}+\angle \mathrm{DCA}+\angle \mathrm{DCB}=180^{\circ}, \\
2 \angle \mathrm{DCA}+2 \angle \mathrm{DCB}=180^{\circ} .
\end{array}
$$

Therefore $\angle \mathrm{ACB}=\angle \mathrm{DCA}+\angle \mathrm{DCB}=90^{\circ}$.