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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.

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It is much simpler than that!

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.

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Also you can note that the circumcircle of $\triangle BDC$ is tangent to $AB$ at $B$, so, by power of $A$, $AB^2=AD\cdot AC$.

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