In $\triangle ABC$, $AB = AC$ and $\angle B = 40^\circ$ . $D$ is a point on $AB$ produced such that $AD = BC$. Join $DC$. Find $\angle DCB$ . 
In $\triangle ABC$, $AB = AC$ and $\measuredangle B = 40^\circ$ . $D$ is a point on $AB$ produced such that $AD = BC$. Join $DC$. Find $\angle DCB$ .

What I Tried: Here is a picture :-

For this problem, I think that I have to find the use of the information that $AD = BC$ , but I cannot find how. I tried angle-chasing, looked for similar triangles and cyclic quadrilaterals, but could not find any which can be that useful.
Can anyone help?
 A: This is a very nice problem - it took me some time but ultimately I found a completely elementary solution. I would be interested to learn about even simpler solutions of this kind, if anyone finds any.
Let $E$ be the point on $AB$ such that $CE$ is the bissector of $\angle ACB$. Next, let $F$ be the point on $BC$ such that $EF$ is the bissector of $\angle BEC$. Now let $G$ be the point on the exterior of line segment $CE$ satisfying $|EF| = |EG|$. Now we observe that $\angle BEC = 180 - 40 - 20 = 120$, and so it follows that $\angle BEF = \angle FEC = \angle AEC = \angle BEG = 60$. From this, two things follow. First, $\triangle FEC$ and $\triangle AEC$ are congruent, and so $|EF| = |EA|$. Second, it follows that $\triangle BEG$ and $\triangle BEF$ are congruent, and so $\angle BGE = \angle BFE = 180 - 40 - 60 = 80$. From this we see that $\triangle BCG$ is equilateral, and so $|BC| = |GC|$. Combining everything we have found, it follows that
$$
|DE| = |DA| - |EA| = |BC| - |EF| = |GC| - |EG| = |CE|.
$$
Thus $\triangle DEC$ is equilateral and so $\angle ECD = (180-120)/2 = 30$. We conclude that $\angle DCB = 30 - 20 = 10$, which concludes the proof.
A: Draw $\angle EAD = \angle ADE = 60^0$. So, $\angle CAE = 40^0$.

$\triangle ACE \cong \triangle BAC$ (by side, side, angle).
As $\triangle ACE$ is isosceles with $AC = CE$, and $\triangle ADE$ is equilateral, with common side $AE$, $CD$ is perpendicular to $AE$.
$\angle BCD = 90^0 - \angle CGF = 90^0 - 80^0 = 10^0 \,$ (as $\angle CGA = 100^0$).
