Find an angle in a $100^\circ$-$40^\circ$-$40^\circ$ triangle without trigonometry (Langley-like solution?) 
In $\triangle ABC$, $A=100^\circ$ and $B=C=40^\circ$. $AB$ is produced to a point $D$ so that $B$ lies between $A$ and $D$ and $AD=BC$. Find $\angle BCD$.

We can find it easily with trigonometry by using sine and cosine rule but since the problem appears to be a variation of Langley's Problem with an $80^\circ$-$80^\circ$-$20^\circ$ triangle, I suppose there must be a geometrical proof of it (or maybe not!).

A possible path 
Consider the Figure below. Draw $DA'$ so that $\angle ADA' = 60^\circ$ and $DA' \cong DA$, and let $E$ be the intersection of $AA'$ with $BC$.



*

*Show that $\triangle AEC$ is isosceles, and thus $EC \cong AE$.

*Demonstrate then that $BE \cong EA'$.

*Use the above information and SAS cryterion to show that $\triangle ABE \cong \triangle A'CE$.

*Therefore $AC\cong CA'$, $ADA'C$ is a kite, and $AA' \perp DC$.

*Conclude that $\angle ADC = 30^\circ$ and $\angle BCD = 10^\circ$. 


Furhter hints

 For 1., note that $\triangle ADA'$ is equilateral, therefore $\angle DAA' = 60^\circ$ and, consequently, $\angle A'AC \cong \angle BCA = 40^\circ$. In order to derive 2., observe that $BE + EC \cong AE + EA'$ (again use the fact that $\triangle ADA'$ is equilateral, plus the original hypothesis $AD \cong BC$), and $EC \cong AE$.

