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Let $\triangle{ABC}$ be an arbitrary triangle. Is it true that for any angles $\alpha, \beta,\gamma\in [0,2\pi]$ with $\alpha+\beta+\gamma=2\pi$ one can find a point $M$ in the plane of the triangle such that: $\angle{(\overline{MB}, \overline{MC})}=\alpha,\ \angle{(\overline{MC}, \overline{MA})}=\beta$ and $\angle{(\overline{MA}, \overline{MB})}=\gamma$ where these angles are oriented?

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Yes, it is true. Let $\Gamma_A$ be the locus of points $P$ for which $\widehat{BPC}=\gamma$, $\Gamma_B$ the locus of points for which $\widehat{CPA}=\beta$. $\Gamma_A$ and $\Gamma_B$ will intersect in a point $M$, other than $C$, for which: $$\widehat{AMB}=2\pi-\widehat{BMC}-\widehat{CMA} = 2\pi-\alpha-\beta=\gamma,$$ done.

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