B2. In the diagram shown, $A B C D$ is a square and point $F$ lies on $B C$. Triangle $D E C$ is equilateral and $E B=E F$.
What is the size of $\angle C E F$ ?
Solution
Since triangle $D E C$ is equilateral, $\angle D C E=60^{\circ}$. Therefore, since $A B C D$ is a square, $\angle E C F=$ $30^{\circ}$. Since $D C=C E$ and $D C=C B$, we have $C E=C B$ and hence triangle $E C B$ is isosceles.

As triangle $E B F$ is isosceles, $\angle C B E=\angle B F E$. Therefore,
$$
\angle B F E=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ} \text { (angles in a triangle sum to } 180^{\circ} \text { ) }
$$
and, consequently,
$$
\angle E F C=105^{\circ} \text { (angles on a straight line sum to } 180^{\circ} \text { ). }
$$

Hence, using the fact that angles in a triangle sum to $180^{\circ}$,
$$
\begin{aligned}
\angle C E F & =180^{\circ}-\angle E F C-\angle E C F \\
& =180^{\circ}-105^{\circ}-30^{\circ} \\
& =45^{\circ} .
\end{aligned}
$$