8. Let $A B C$ be a triangle with $A B=A C$ and $\angle B A C=20^{\circ}$. Let $D$ be the point on the side $A B$ such that $\angle B C D=70^{\circ}$. Let $E$ be the point on the side $A C$ such that $\angle C B E=60^{\circ}$. Determine the value of the angle $\angle C D E$.

Answer: $\angle C D E=20^{\circ}$.
Solution 1: Define $E^{\prime}$ on the side $A B$ such that $\angle B C E^{\prime}=60^{\circ}$, and $X$ the intersection of $B E$ and $C E^{\prime}$. Straightforward angle chasing gives that $\Delta E^{\prime} E X$ is equilateral, as it has angles of $60^{\circ}$. Moreover, $\triangle A E^{\prime} C$ is isosceles at $E^{\prime}$, as $\angle E^{\prime} A C=20^{\circ}=\angle E^{\prime} C A$. Consider the reflection preserving this triangle, sending $E^{\prime}$ to itself, and exchanging $A$ and $C$. Then $D$ and $X$ are also exchanged, as $\angle E^{\prime} C D=10^{\circ}=\angle E^{\prime} A X$, in particular $E^{\prime} D=E^{\prime} X$. As $\Delta E^{\prime} E X$ is equilateral, we can now deduce that $E^{\prime} D=E^{\prime} X=E^{\prime} E$, so that $\triangle D E^{\prime} E$ is isosceles at $E^{\prime}$. The rest is angle chasing:

and

$$
\angle E^{\prime} D E=\frac{180^{\circ}-\angle D E^{\prime} E}{2}=\frac{180^{\circ}-80^{\circ}}{2}=50^{\circ}
$$

$$
\angle C D E=\angle E^{\prime} D E-\angle E^{\prime} D C=50^{\circ}-30^{\circ}=20^{\circ}
$$