# Elementary euclidean geometry problem

The problem states:

Consider a triangle $$\Delta{ABC}$$ in which $$AC\gt AB$$. A half line with origin in B cuts AC in D such that the angles $$\angle ABD$$ and $$\angle BCD$$ are equal. Prove that $$(AB)^2=AC\cdot AD$$

Can anyone give hints on how to tackle this problem? I have not done mathematical proofs for quite some time. I tried using Stewart's theorem, which in this case yields:

$$(BA)^2 \cdot DC +(BD)^2 \cdot CA +(BC)^2 \cdot AD + (AD)\cdot(DC)\cdot(CA)=0$$

Then I tried using the angle bisector theorem to get rid of some terms ($$\frac{AD}{DC}=\frac{BA}{BC}$$) but I still cannot prove the problem. Any help will be very much appreciated.

Note that $$\angle BCD$$ is just another name for $$\angle BCA$$, so you have $$\triangle ACB\sim\triangle ABD$$ since $$\angle A$$ is common and one other angle $$\angle ACB=\angle ABD$$. Thus $$\frac{AB}{AC}=\frac{AD}{AB}$$ which rearranges to what you want.
Also you can note that the circumcircle of $$\triangle BDC$$ is tangent to $$AB$$ at $$B$$, so, by power of $$A$$, $$AB^2=AD\cdot AC$$.