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**Unbeatable0** A figure is added in the attachment to make the proof easier to read.

On the given triangle $\displaystyle \triangle ABC$ build a regular $\displaystyle 18$-gon such that $\displaystyle A$ is the center of it

(which is possible due to the fact that $\displaystyle AB=AC$, $\displaystyle \angle BAC = 20^{\circ}$).

Now, we **are not** assuming $\displaystyle AD=BC$. We'll come back to it later.

Continue AB until it gets to another vertex, call it $\displaystyle J$, of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line $\displaystyle BS$ as shown in the attached figure. Next, draw $\displaystyle JT$ as shown, and call $\displaystyle M$ the intersection of $\displaystyle JT$ with $\displaystyle BS$. All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle $\displaystyle \triangle JMS$ it follows that $\displaystyle [JM]=2[JS]$, and in triangle $\displaystyle \triangle JMB$ we have $\displaystyle [JM]=2[AD]$. Therefore $\displaystyle [JS]=[AD]$, but $\displaystyle [JS]=[BC]$, so $\displaystyle [AD]=[BC]$. Next notice that $\displaystyle \angle JBS=10^\circ$ (again, by considering the circumscribing circle).

We have thus proven that the conditions $\displaystyle \angle JBS=10^\circ$ and $\displaystyle [AD]=[BC]$ are obtained simultaneously, from which the result follows.