Exercise 12. Let $A B C$ be an isosceles triangle at $A$ such that $\widehat{C B A}=61^{\circ}$. Let $E$ be the point, other than $A$, located on the circumcircle of $A B C$ such that $E B=E C$. Let $D$ be the point other than $A$ such that $D B=D C=A B$.

Determine the value in degrees of the angle $\widehat{B E D}$.
Solution to Exercise 12: We have $\widehat{B A C}=180^{\circ}-2 \widehat{C B A}=58^{\circ}$.

$D$ is the symmetric point of $A$ with respect to the line $(A B)$. We have $\widehat{B E D}=180^{\circ}-$ $\widehat{A E B}=180^{\circ}-\left(90^{\circ}-\widehat{B A E}\right)=90^{\circ}+\frac{1}{2} \widehat{B A C}=90+29=119^{\circ}$.