Marinada '23

[lang = hr] Neka je $ABC$ trokut u kome je $\angle BAC = 60^{\circ}$. Neka su $D,E$ točke na $BC$ i $AC$ redom takve da je $AD$ simetrala $\angle BAC$ i $BE$ je simetrala $\angle ABC$. Ako vrijedi $AB + BD = AE + EB$, koliko je $\angle ABC - \angle ACB$? Rješenje zaokruži na dvije decimale. [/lang] [lang = en] Let $ABC$ be a triangle in which $\angle BAC = 60^{\circ}$. Let $D,E$ be points on $BC$ and $AC$ respectively such that $AD$ is the bisector of $\angle BAC$ and $BE$ is the bisector of $\angle ABC$. If $AB + BD = AE + EB$ holds, what is $\angle ABC - \angle ACB$? Round the solution to two decimal places. [/lang]