A problem from INMO

 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.

Scroll down for a solution

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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.