<s>[INST] How to solve this complicated trigonometric equation for the solution of this triangle. 
In $\triangle ABC$, if $AB=AC$ and internal bisector of angle $B$ meets $AC$ at $D$ such that $BD+AD=BC=4$, then what is $R$ (circumradius)?


My Approach:- I first drew the diagram and considered $\angle ABD=\angle DBC=\theta$ and as $AB=AC$, $\angle C=2\theta$.  Therefore $\angle A=180-4\theta$. Also as $AE$ is the angular bisector and $AB=AC$, then $BE=EC=2.$ Now applying the sine theorem to $\triangle ADB$ and  $ \triangle BDC$ gives $$\frac{BD}{\sin {(180-4\theta)}}=\frac{AD}{\sin \theta}$$
$$\frac{BC}{\sin(180-3\theta)}=\frac{BD}{\sin 2\theta}$$
Now we know that $BC=4$ and then solving both the equations by substituting in $BD+AD=4$, we get $$\sin 2\theta .\sin4\theta+\sin2\theta.\sin\theta=\sin3\theta.\sin4\theta$$
$$\sin4\theta+\sin\theta=\frac{\sin3\theta.\sin4\theta}{\sin2\theta}$$
Now I have no clue on how to proceed further from here. Though I tried solving the whole equation into one variable ($\sin\theta$), but it's getting very troublesome as power of $4$ occurs. Can anyone please help further or else if there is any alternative method to solving this problem more efficiently or quickly?
Thank You
 [/INST] The values are such that equations do not simplify. Nonetheless, here is an alternate approach.
Say $AB = AC = x$ and we know $BC = 4$,
By angle bisector theorem,
$\cfrac{4}{x} = \cfrac{x-AD}{AD} \implies AD = \cfrac{x^2}{4+x}$
Now by angle bisector length formula,
$BD^2 = \cfrac{4x}{(4+x)^2} [(4+x)^2 - x^2] = \cfrac{32x(x+2)}{(4+x)^2}$
Now, $BD + AD = 4 = \cfrac{x^2}{4+x} + \cfrac{\sqrt{32x(x+2)}}{4+x}$
$16+4x-x^2 = \sqrt{32x(x+2)}$
Solving using WolframAlpha, the only valid solution is $x \approx 2.61$
Now to find circumradius, use $R = \cfrac{abc}{4 \triangle} = \cfrac{4 x^2}{8 \sqrt{x^2 - 4}}$
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