<s>[INST] $AB=AC$, $BD$ 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
 [/INST] 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.
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