Problem on angles In triangle ABC, AB = AC and angle BAC = 20. A point D lies on the side AB and AD = BC. Find angle BCD?
My construction involved drawing a line parallel to BC intersecting AC at E. However, I am still unable to link this to the known angles! 
Hint please!
 A: 
Let $\angle BCD=\theta$
If you apply sine rule in the $\Delta ADC$ you'll get :
$$\frac{AD}{\sin(80^\circ-\theta)}=\frac{DC}{\sin20^\circ}$$
And doing the same in the $\Delta BDC$ you'll get:
$$\frac{BC}{\sin(100^\circ-\theta)}=\frac{DC}{\sin80^\circ}$$
Dividing these two relations you get:
$$\frac{\sin(100^\circ-\theta)}{\sin(80^\circ-\theta)}=\frac{\sin80^\circ}{\sin20^\circ}$$
EDIT:I had time so I decided to add how I got the value of $\theta=70^\circ$
$$\sin20^\circ\sin(100^\circ-\theta)=\sin80^\circ\sin(80^\circ-\theta)$$
$$\sin20^\circ\sin(80^\circ+\theta)=\sin80^\circ\sin(80^\circ-\theta)$$
$$\sin20^\circ\sin80^\circ\cos\theta+\sin20^\circ\cos80^\circ\sin\theta=\sin^280^\circ\cos\theta-\sin80^\circ\cos80^\circ\sin\theta$$
Rearranging,
$$\tan\theta=\tan80^\circ{\sin80^\circ-\sin20^\circ \over \sin80^\circ+\sin20^\circ}$$
$$\tan\theta=\tan80^\circ{\cos50^\circ \sin30^\circ \over \sin50^\circ \cos30^\circ }$$
$$\tan\theta=\tan80^\circ\tan40^\circ\tan30^\circ$$
$$\tan\theta={\tan80^\circ\tan40^\circ\tan30^\circ\tan20^\circ\over \tan20^\circ}$$
Now just apply this identity with $x=20^\circ$:
$$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=tan3x$$
$$\tan\theta={1\over \tan20^\circ}=\tan70^\circ$$
