3. Inside triangle $ABC$, a point $P$ is chosen such that $AP=BP$ and $CP=AC$. Find $\angle CBP$, given that $\angle BAC = 2 \angle ABC$.

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Here is the translation of the provided text, maintaining the original formatting and structure.
Answer: $30^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-19.jpg?height=500&width=1312&top_left_y=1138&top_left_x=450)

Solution 1. Draw the perpendicular bisector of side $A B$. Obviously, it will pass through point $P$. Let $C^{\prime}$ be the point symmetric to $C$ with respect to this perpendicular (see the left figure). By symmetry, $\angle C^{\prime} B A = \angle C A B = 2 \angle A B C$, hence $\angle C^{\prime} B C = \angle A B C$. On the other hand, lines $C C^{\prime}$ and $A B$ are parallel, so $\angle A B C = \angle B C C^{\prime}$. Therefore, $\angle B C C^{\prime} = \angle C^{\prime} B C$, triangle $B C^{\prime} C$ is isosceles, and $B C^{\prime} = C^{\prime} C$. By symmetry, $B C^{\prime} = A C$ and $C P = C^{\prime} P$, and by the condition $A C = C P$. Thus,

$$
C^{\prime} P = C P = A C = B C^{\prime} = C^{\prime} C
$$

and triangle $P C C^{\prime}$ is equilateral. Therefore, $\angle C C^{\prime} P = 60^{\circ}$. Note that

$$
180^{\circ} - 2 \angle P B C^{\prime} = \angle P C^{\prime} B = \angle B C^{\prime} C - 60^{\circ} = 120^{\circ} - 2 \angle C B C^{\prime}
$$

Therefore,

$$
\angle C B P = \angle P B C^{\prime} - \angle C B C^{\prime} = \frac{1}{2} \left(180^{\circ} - 120^{\circ}\right) = 30^{\circ}.
$$

Solution 2. Let $\alpha = \angle P A B, \beta = \angle P A C, \gamma = \angle P B C, \varphi = \alpha + \gamma$ (see the right figure). By the condition, $\angle P B A = \alpha, \angle C P A = \beta$, and

$$
\alpha + \beta = 2 \varphi = 2(\alpha + \gamma) \Longleftrightarrow 2 \gamma = \beta - \alpha
$$

By the Law of Sines,

$$
\frac{A C}{\sin \varphi} = \frac{A B}{\sin \left(180^{\circ} - 3 \varphi\right)} \Longleftrightarrow \frac{\sin 3 \varphi}{\sin \varphi} = \frac{A B}{A C}
$$

Note that

$$
A B = 2 A P \cos \alpha = 4 A C \cos \alpha \cos \beta \Longleftrightarrow \frac{A B}{A C} = 2(\cos (\alpha + \beta) + \cos (\beta - \alpha)) = 2 \cos 2 \varphi + 2 \cos 2 \gamma
$$

Moreover,

$$
\frac{\sin 3 \varphi}{\sin \varphi} = 3 - 4 \sin^2 \varphi = 2 \cos 2 \varphi + 1
$$

Therefore,

$$
2 \cos 2 \varphi + 2 \cos 2 \gamma = 2 \cos 2 \varphi + 1 \Longleftrightarrow \cos 2 \gamma = \frac{1}{2} \Longleftrightarrow 2 \gamma = 60^{\circ} \Longleftrightarrow \gamma = 30^{\circ}
$$