9. In the triangle $\triangle A B C, A B=A C, \angle A B C=40^{\circ}$ and the point $D$ is on $A C$ such that $B D$ is the angle bisector of $\angle A B C$. If $B D$ is extended to the point $E$ such that $D E=A D$, find $\angle E C A$.
(A) $20^{\circ}$
(B) $30^{\circ}$
(C) $40^{\circ}$
(D) $45^{\circ}$
(E) $50^{\circ}$
9. Answer: (C)
Construct a point $F$ on $B C$ such that $B F=B A$. Since $\angle A B D=\angle F B D, \triangle A B D$ is congruent to $\triangle F B D$.
Thus $D F=D A=D E$ and $\angle F D B=\angle A D B=60^{\circ}$. We also have $\angle E D C=$ $\angle A D B=60^{\circ}$, which implies that $\angle F D C=60^{\circ}$ and $\triangle C F D$ is congruent to $\triangle C E D$. In conclusion, $\angle E C D=\angle F C D=40^{\circ}$.