In a triangle $ABC$ (sides $a,b,c$ opposite $A,B,C$), angle $A$ is twice $B$.
Show that $$a^2 = b(b+c)$$
Try not to use trigonometry if possible.
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Let AD be the angular bisector of A, D lying on BC (might help to draw a figure).
Then by angular bisector theorem
$$BD = \frac{ac}{b+c}, DC = \frac{ab}{b+c}$$
BAD is isosceles, with $AD = BD$. Also triangle $ADC$ is similar to triangle $BAC$.
$AD/AB = DC/AC$ gives the result.
There are non-trigonometric proofs of the angular bisector theorem. For eg, prove for right angled triangles and use affine transform etc.
