Marinada '22

[lang=hr] Neka je $ABC$ trokut s opisanom kružnicom $k$. Točka $U$ je središte luka $AB$ kružnice $k$ koji ne sadrži točku $C$, točka $V$ je središte luka $BC$ kružnice $k$ koji ne sadrži točku $A$. Točka $S$ je na $AC$ takva da je $BS$ simetrala kuta $\angle ABC$. Poznato nam je da vrijedi $\angle ABC=2\angle ACB$ i $\angle USV=90^{\circ}$. Odredi $|\angle ABC - \angle BAC|$. (rješenje zapišite u stupnjevima bez mjerne jedinice npr. "45") [/lang] [lang=en] Let $ABC$ be a triangle with a circumscribed circle $k$. Point $U$ is the midpoint of circular arc $AB$ of circle $k$ that doesn't contain point $C$, point $V$ is the midpoint of circular arc $BC$ of circle $k$ that doesn't cointain point $A$. Choose point $S$ on $AC$ such that $BS$ bisects angle $\angle ABC$. Given that $\angle ABC=2\angle ACB$ and $\angle USV=90^{\circ}$ determine $|\angle ABC - \angle BAC|$. (Write the solution in degrees but without measuring unit, e.g. "45".) [/lang]