5. Side $A B$ of triangle $A B C$ is greater than side $B C$, and angle $B$ is $40^{\circ}$. A point $P$ is taken on side $A B$ such that $B P = B C$. The bisector $B M$ intersects the circumcircle of triangle $A B C$ at point $T$. Find the angle $M P T$.
Answer: $20^{\circ}$.

Solution. (Fig. 3.) In the quadrilateral $A P M T$, the angle at vertex $A$ is measured by half the arc $T C B$. Triangles $P M B$ and $C M B$ are equal by two sides and the angle between them, so $\angle P M B = \angle C M B = \angle A M T$. The angle $\angle A M T$ is measured by half the sum of the arcs $A T$ and $C B$, and:

$$
\frac{1}{2}(\overparen{A T} + \overparen{C B}) = \frac{1}{2}(\overparen{T C} + \overparen{C B}) = \frac{1}{2} \overparen{T C B}
$$

Thus, $\angle C M B = \angle B A T$. Therefore, $\angle P M B = \angle B A T$ and $\angle B A T + \angle P M T = \angle B A T + (180^{\circ} - \angle P M B) = 180^{\circ}$. Consequently, the sum of the opposite angles of the quadrilateral $A P M T$ is $180^{\circ}$, meaning $A P M T$ is cyclic. By the property of inscribed angles,

![](https://cdn.mathpix.com/cropped/2024_05_06_4c1598eaf6f0860458ecg-2.jpg?height=496&width=467&top_left_y=1145&top_left_x=1512)

Fig. 3 $\angle M P T = \angle M A T = \angle C A T = \angle T B C = \frac{1}{2} \angle A B C = 20^{\circ}$.

Criteria. The equality of angles $P M B$ and $C M B$ (or $A M T$) is noted - 1 point. It is proven that the angle $A M T$ or $C M B$ is equal to the angle $B A T$ - 2 points. It is proven that the quadrilateral $A P M T$ is cyclic - 5 points. The criteria are not cumulative. Full solution - 7 points.