Utilizing the area ratio formula (please see %%HREF%%Geometry Techinque%%http://www.mathallstar.org/Home/Home/35-books/94-book-geometry-techniques%%): $$S_{\triangle{ABC}}:S_{\triangle{ADC}}=BE:BD=1$$ On the other hand: $$S_{\triangle{ABC}} = \frac{1}{2}\cdot AB \cdot BC\sin\angle{ABC}$$ $$S_{\triangle{ADC}} = \frac{1}{2}\cdot AD \cdot DC\sin\angle{ADC}$$ Because $ABCD$ are concyclic: $$\angle{ABC} + \angle{ADC}=180^\circ\implies \sin\angle{ABC}=\sin\angle{ADC}$$ Therefore: $AB\cdot BC = AD\cdot DC$. Next, by the law of cosine: $$AC^2 = AB^2 + BC^2 - 2\cdot AB\cdot BC\cdot\cos\angle{ABC}$$ $$AC^2 = CD^2 + DA^2 - 2\cdot CD\cdot DA\cdot\cos\angle{ADC}$$ Note that $\cos\angle{ABC} = -\cos\angle{ADC}$, adding the above two equations leads to the desired result.