> 
> In $\triangle ABC$, the points $D$ and $E$ are located on $\overline{BC}$ such that $\angle BAD \cong \angle CAE$. Prove that 
>  $$\frac{AB^2}{AC^2}=\frac{BD\cdot BE}{CD\cdot CE}$$
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> 
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Since the condition to prove looks a bit like the power of a point theorem, I tried constructing the circumcircle of $\triangle ADE$. The condition is easy to prove assuming that $\overline{AB}$ and $\overline{AC}$ are tangents to the circumcircle, but how do I prove that they *are* tangents? Since I haven't yet used the fact that $\angle BAD$ and $\angle CAE$ are equal, can it be proven using that condition?