Find the angle in an isosceles triangle Let triangle $\Delta ABC$ have $AB=AC$. Then we draw the angle bisector from $B$ to $AC$ intersecting at $D$. Find the angle $\angle BAC$ if $BC=AD+BD$.
My attempts:
I know that the answer is 100° but I couldn't prove that if you extended $AD$ to a point $E$ so it is equal to $BC$, then the angle $\angle DCE$ is the same as $\angle ACB$. 

In the standard notation we obtain:
$$\frac{AD}{DC}=\frac{AB}{BC}=\frac{c}{a}$$ and $$AD+DC=AC=b,$$ which gives
$$AD=\frac{bc}{a+c},$$ $$DC=\frac{ab}{a+c}$$ and
$$BD^2=AB\cdot BC-AD\cdot DC=ac-\frac{b^2ac}{(a+c)^2}=ab-\frac{ab^3}{(a+b)^2},$$
which gives $$BD=\frac{a\sqrt{b(a+2b)}}{a+b}.$$
Id est, by the given we obtain:
$$a=\frac{b^2}{a+b}+\frac{a\sqrt{b(a+2b)}}{a+b}$$ or
$$a\sqrt{b(a+2b)}=a^2+ab-b^2.$$
Now, $\sin\frac{\alpha}{2}=\frac{\frac{a}{2}}{b}=\frac{a}{2b}.$
Let $\sin\frac{\alpha}{2}=x.$
Thus, $a=2xb$ and we obtain:
$$2x\sqrt{2(x+1)}=4x^2+2x-1$$ or
$$8x^2(x+1)=(4x^2+2x-1)^2,$$ where $x>0$ and $4x^2+2x-1>0.$
We obtain:
$$16x^4+8x^3-12x^2-4x+1=0$$ or
$$8x^3(2x+1)-12x^2-6x+2x+1=0$$ or
$$(2x+1)(8x^3-6x+1)=0$$ or
$$3x-4x^3=\frac{1}{2}$$ or
$$\sin\frac{3\alpha}{2}=\frac{1}{2},$$ which gives
$$\frac{3\alpha}{2}=30^{\circ},$$ which is impossible because $4x^2+2x-1>0,$ or
$$\frac{3\alpha}{2}=150^{\circ},$$ which gives $\alpha=100^{\circ}$ and $\beta=\gamma=40^{\circ}.$
The fact that $$BD^2=AB\cdot BD-AD\cdot DC$$ we can prove by the following reasoning.
Let $\Phi$ be a circumcircle of $\Delta ABC$ and $AD\cap\Phi=\{A,E\}$.
Thus, $\measuredangle BAC=\measuredangle BEC$ and $\measuredangle ABD=\measuredangle EBD,$ which gives $\Delta ABD\sim\Delta EBC.$
Hence, $$\frac{AB}{BE}=\frac{BD}{BC}$$ or
$$AB\cdot BC=BD\cdot BE$$ or
$$AB\cdot BC=BD(BD+DE)$$ or
$$BD^2=AB\cdot BC-BD\cdot DE$$ or
$$BD^2=AB\cdot BC-AD\cdot DC.$$
