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Inspired by a recent post, consider the following problem. You are given the four sides lengths of a quadrilateral $ABCD$. Let these sides be $a = AB , b = BC, c = CD, d = DA $. I want to determine the angles $ \phi = \angle DAB $ and $\theta = \angle ABC $, that will make this quadrilateral concyclic (i.e. having its vertices on a single circle). That is the task. As a hint, from this Wikipedia page we know that such a quadrilateral maximizes the area $[ABCD]$ of the quadrilateral.

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COMMENT.-It is easy if we take into account that the opposite to the angle in a vertex must be its suplement. In fact $$\overline{DB}^2=a^2+d^2-2ad\cos(\phi)=b^2+c^2-2bc\cos(\pi-\phi)=b^2+c^2+2bc\cos(\phi)$$ so $$\cos(\phi)=\frac{a^2+d^2-(b^2+c^2)}{2(ad+bc)}$$Similarly with the other angle.

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  • $\begingroup$ Oh yeah, that's true. How could I miss that? $\endgroup$ Feb 26 at 15:59

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