Coordinate Geometry?

Construct $AD$ , $D$ on $BC$ , such that $AD=DB\Rightarrow \angle BAD=\angle CAD=\angle ABD$ . Let $AD=d=DB$ , so $CD=a-d$ .

Now, we use Stewart's theorem and angle bisector theorem to derive some interesting results:

$b^2d+c^2(a-d)=a(a-d)d+d^2a$ $ac-cd=bd \Rightarrow ac=bd+cd$

Let us expand out the first equation first. $b^2d+c^2a-c^2d=a^2d-ad^2+ad^2\\a^2d=b^2d-c^2d+c^2$

Now, sub in the second equation. $a^2d=b^2d-c^2d+(bd+cd)c\\a^2d=b^2d+bcd\\a^2=b(b+c)$

Hence, $a^2-b(b+c)=0$

This is a piece of cake. Let the triangle be a 30-60-90 one.

So <B = 30, <A = 60, <C = 90

a = sqrt(3) b = 1 c = 2

a^2 - b(b + c) = 3 - 1(1 + 2) = 0 --> Answer

$A=2B\ \ \ \ \ \ \ \ \ Sin2B=2*CosB*SinB\\ SinC=Sin(180 - A - B)=Sin(A+B)=Sin3B=SinB(3 - 4Sin^2B)=SinB(4Cos^2B - 1) \\ Using\ Sin\ Law:-\\ \dfrac{a}{Sin2B}=\dfrac{b}{SinB}=\dfrac{c}{SinC}\ \ \ \implies\ \dfrac{a}{2SinB*CosB}=\dfrac{b}{SinB}=\dfrac{c}{SinB(4Cos^2B - 1)}\\ \therefore\ \ a^2=(2CosB)^2*b^2,\ \ \ c=(4Cos^2B - 1)*b.\\ \therefore\ \ a^2 - b(b+c)=4Cos^2B*b^2 - b\{b+(4Cos^2B - 1)b\} = 0.$

Solutions already given here really nice!I used an isosceles right angled triangle right angled at A with side lengths b=c=1.Since the only relation we need to take care of is angle A is twice angle B,made me go for this particular example to get the answer.