1. In the diagram, $B D$ and $A E$ intersect at $C$ and $A B=B C$. Also, $\angle A B C=40^{\circ}$ and $\angle D C E=x^{\circ}$. What is the value of $x$ ?
1. Solution 1
Since opposite angles are equal, $\angle A C B=\angle D C E=x^{\circ}$.
In $\triangle A B C, A B=B C$, and so $\angle C A B=\angle A C B=x^{\circ}$.

Therefore,
$$
\begin{aligned}
\angle A C B+\angle C A B+\angle A B C & =180^{\circ} \\
x^{\circ}+x^{\circ}+40^{\circ} & =180^{\circ} \\
2 x & =140
\end{aligned}
$$
and so $x=70$.
Solution 2
In $\triangle A B C, A B=B C$, and so $\angle C A B=\angle A C B$.
Since the sum of the angles in $\triangle A B C$ is $180^{\circ}$, then
$$
\begin{aligned}
\angle A C B+\angle C A B+\angle A B C & =180^{\circ} \\
2 \angle A C B+40^{\circ} & =180^{\circ} \\
2 \angle A C B & =140^{\circ} \\
\angle A C B & =70^{\circ}
\end{aligned}
$$

Since $\angle D C E$ and $\angle A C B$ are opposite angles, then $x^{\circ}=70^{\circ}$ and so $x=70$.
ANSWER: $x=70$