$AB=AC$, $BD$ is the angle bisector of $\angle B$ , find $\angle A$ Let $ABC$ be an triangle, $AB=AC$. $BD$ is the angle bisector of $\angle B$, $BD$ intersect $AC$ at point $D$, and $AD=BC+BD$. 
show that: $\measuredangle BAC=20^\circ$

Well, If $\measuredangle BAC=20^\circ$, I can prove that  $AD=BC+BD$. 
But, if $AD=BC+BD$, How to show $\measuredangle BAC=20^\circ$?
Thanks a lot
 A: Let $F$ be the point such that $DF||BC$.
Clearly $BCDF$ is cyclic and since $BD$ is the angular bisector of $\angle CBF$, this implies $CD=DF$.
Let $E$ be the point on $BC$ such that $CE=DE$.
Note that triangles $ECD$ and $AFD$ are similar and since $CD=DF$, they are congruent.
Hence $CE=AD$ and the condition implies $BC+BD=AD=CE=BC+BE$ i.e. $BD=BE$.
So triangle $BDE$ is isosceles and we can now finish this problem off with some angle chasing.
Denote $\angle BAC=4\alpha$.
Then we have $$90^\circ-2\alpha=\measuredangle EDC=\measuredangle EDB+\measuredangle BDC$$$$=\measuredangle BED+180^\circ-\measuredangle DCB-\measuredangle CBD$$$$=4\alpha+180^\circ-(90^\circ-2\alpha)-(45^\circ-\alpha)=45^\circ+7\alpha$$Thus $\alpha=5^\circ$ and hence
$\measuredangle BAC=4\alpha=20^\circ$.

A: By Stewart's theorem we have:
$$ BD^2 = \frac{ab}{(a+b)^2}\left((a+b)^2-c^2\right)=\frac{a^2 b}{(a+b)^2}(a+2b) $$
and by the bisector theorem we have $AD=\frac{b^2}{a+b}$, so $BD^2=(AD-a)^2$ leads to:
$$ a^2 b(a+2b)= (a^2+ab-b^2)^2 $$
so by assuming $b=1$, $a=2\sin\frac{\widehat{BAC}}{2}$ is a root of:
$$ p(x) = x^4+x^3-3x^2-2x+1 = (x+1)(x^3-3x+1). $$
In order to prove $\widehat{BAC}=20^\circ$, we just need to prove that $\sin\frac{\pi}{18}=\cos\frac{4\pi}{9}$ is a root of:
$$ q(x) = 8x^3-6x+1 = 2\cdot T_3(x)+1, $$
that is trivial by using Chebyshev polynomials of the first kind.
