## Prvo rješenje.

Neka je $c=|A B|=|C D|$, te $\alpha=\varangle B A C$ i $\gamma=\varangle A C B$.

Tada je $\varangle C A D=\varangle B A D=\frac{\alpha}{2}$ te $\varangle B D A=\frac{\alpha}{2}+\gamma$. 1 bod

Primjenom poučka o sinusima na trokute $A D C$ i $A B D$ dobivamo

$$
\frac{\sin \gamma}{\sin \frac{\alpha}{2}}=\frac{|A D|}{c}, \quad \frac{\sin 2 \gamma}{\sin \left(\frac{\alpha}{2}+\gamma\right)}=\frac{|A D|}{c}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-16.jpg?height=506&width=765&top_left_y=181&top_left_x=583)

Dalje imamo

$$
\begin{aligned}
\sin \gamma \sin \left(\frac{\alpha}{2}+\gamma\right) & =\sin 2 \gamma \sin \frac{\alpha}{2}, & & 1 \text { bod } \\
\sin \gamma \sin \left(\frac{\alpha}{2}+\gamma\right) & =2 \sin \gamma \cos \gamma \sin \frac{\alpha}{2}, & & 1 \text { bod } \\
\sin \left(\frac{\alpha}{2}+\gamma\right) & =2 \cos \gamma \sin \frac{\alpha}{2}, & & \\
\sin \frac{\alpha}{2} \cos \gamma+\cos \frac{\alpha}{2} \sin \gamma & =2 \cos \gamma \sin \frac{\alpha}{2}, & & 2 \text { boda } \\
\cos \frac{\alpha}{2} \sin \gamma & =\sin \frac{\alpha}{2} \cos \gamma, & & 1 \mathrm{bod}
\end{aligned}
$$

Budući da su $\alpha$ i $\gamma$ kutovi trokuta, jedina mogućnost je $\gamma=\frac{\alpha}{2}$.

Vrijedi $\alpha+3 \gamma=180^{\circ}$, pa je $\alpha=\frac{2}{5} \cdot 180^{\circ}=72^{\circ}$.


## Zadatak A-3.6.

U trokutu $A B C$ vrijedi $\varangle A B C=2 \varangle B C A$. Simetrala kuta $\varangle B A C$ siječe stranicu $\overline{B C}$ u točki $D$ tako da je $|A B|=|C D|$. Odredi $\varangle C A B$.

