## Rješenje.

![](https://cdn.mathpix.com/cropped/2024_05_30_d6d90a155ebaa296cbbeg-01.jpg?height=366&width=1016&top_left_y=1690&top_left_x=520)

Neka je $\varangle D B C=2 \varangle A B D=2 x$ i $\varangle D C B=2 \varangle A C D=2 y$.

Promatrajući kutove trokuta $A B C$ dolazimo do:

$$
\begin{gathered}
\varangle B A C+\varangle C B A+\varangle A C B=180^{\circ} \\
120^{\circ}+3 x+3 y=180^{\circ} \\
x+y=20^{\circ}
\end{gathered}
$$

stoga iz trokuta $D B C$ slijedi

$$
\begin{aligned}
\varangle B D C & =180^{\circ}-\varangle C B D-\varangle D C B \\
& =180^{\circ}-2 x-2 y \\
& =180^{\circ}-40^{\circ}=140^{\circ} .
\end{aligned}
$$


## Zadatak A-1.2. (4 boda)

U trokutu $A B C$ vrijedi $\varangle B A C=120^{\circ}$. Točka $D$ nalazi se unutar trokuta tako da vrijedi $\varangle D B C=2 \varangle A B D$ i $\varangle D C B=2 \varangle A C D$. Izračunaj mjeru kuta $\varangle B D C$.

