Dgdrxrt

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:
$$fracADDC=fracABBC=fracca$$ and $$AD+DC=AC=b,$$ which gives
$$AD=fracbca+c,$$ $$DC=fracaba+c$$ and
$$BD^2=ABcdot BC-ADcdot DC=ac-fracb^2ac(a+c)^2=ab-fracab^3(a+b)^2,$$
which gives $$BD=fracasqrtb(a+2b)a+b.$$
Id est, by the given we obtain:
$$a=fracb^2a+b+fracasqrtb(a+2b)a+b$$ or
$$asqrtb(a+2b)=a^2+ab-b^2.$$
Now, $sinfracalpha2=fracfraca2b=fraca2b.$

Let $sinfracalpha2=x.$

Thus, $a=2xb$ and we obtain:
$$2xsqrt2(x+1)=4x^2+2x-1$$ 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
$$3x-4x^3=frac12$$ or
$$sinfrac3alpha2=frac12,$$ which gives
$$frac3alpha2=30^circ,$$ which is impossible because $4x^2+2x-1>0,$ or
$$frac3alpha2=150^circ,$$ which gives $alpha=100^circ$ and $beta=gamma=40^circ.$

The fact that $$BD^2=ABcdot BD-ADcdot DC$$ we can prove by the following reasoning.

Let $Phi$ be a circumcircle of $Delta ABC$ and $ADcapPhi=A,E$.

Thus, $measuredangle BAC=measuredangle BEC$ and $measuredangle ABD=measuredangle EBD,$ which gives $Delta ABDsimDelta EBC.$

Hence, $$fracABBE=fracBDBC$$ or
$$ABcdot BC=BDcdot BE$$ or
$$ABcdot BC=BD(BD+DE)$$ or
$$BD^2=ABcdot BC-BDcdot DE$$ or
$$$BD^2=ABcdot BC-ADcdot DC.$$