Suggestion: Show that triangle BEC is isosceles.

Facts that Help: The sum of the measures of the interior angles of a triangle is equal to $180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_64d7a736a938f072abd8g-4.jpg?height=454&width=402&top_left_y=1903&top_left_x=193)

Inside a triangle $ABC$, a point $E$ is taken such that $AE = BE$ and $AB = EC$. If $A \hat{B E} = \alpha = E \hat{C A}$, $E \hat{A} C = 2 \alpha$ and $E \hat{B C} = 5 \alpha$, determine $\alpha$.

##
Knowing that the sum of the internal angles of a triangle is $180^{\circ}$, we obtain

$$
\left\{\begin{array}{l}
A \hat{E} B=180^{\circ}-(\alpha+\alpha)=180^{\circ}-2 \alpha \\
A \hat{E} C=180^{\circ}-(\alpha+2 \alpha)=180^{\circ}-3 \alpha
\end{array}\right.
$$

Thus, we have

$$
\begin{aligned}
\mathrm{C} \hat{\mathrm{E}} \mathrm{B} & =360^{\circ}-(\mathrm{A} \hat{\mathrm{E}} \mathrm{C}+\mathrm{A} \hat{\mathrm{E}} \mathrm{B})= \\
& =360^{\circ}-\left(180^{\circ}-3 \alpha+180^{\circ}-2 \alpha\right)=5 \alpha
\end{aligned}
$$

Since the angle $E \hat{B C}$ also measures $5 \alpha$, it follows that triangle $B E C$ is isosceles. Thus, $A B=C E=B C$, that is, triangle $A B C$ is also isosceles.

Therefore, $\mathrm{BC} A=B \hat{A} C=3 \alpha$ and $B \hat{C} A+C \hat{A} B+A \hat{B C}=180^{\circ}$, that is, $3 \alpha+$ $3 \alpha+6 \alpha=180^{\circ}$, which results in $\alpha=15^{\circ}$.