Find the measure of angle $B \widehat{A} D$, knowing that $D \widehat{A} C=39^{\circ}, A B=A C$ and $A D=B D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-032.jpg?height=240&width=463&top_left_y=2441&top_left_x=1365)
Given $A B=A C$, the triangle $\triangle A B C$ is isosceles, so $A \widehat{B} C=A \widehat{C} B$. Since $A D=B D$, the triangle $\triangle A B D$ is also isosceles, so $A \widehat{B} D=B \widehat{A} D$. Therefore,

$$
A \widehat{C} B=A \widehat{B} C=A \widehat{B} D=B \widehat{A} D
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-098.jpg?height=239&width=457&top_left_y=1174&top_left_x=1416)

In the figure, these three equal angles are represented by the letter $\alpha$. The internal angles of $\triangle A B C$ are $\alpha+39^{\circ}, \alpha$ and $\alpha$. Therefore, $\alpha+39^{\circ}+\alpha+\alpha=180^{\circ}$, which means $3 \alpha=180^{\circ}-39^{\circ}=$ $141^{\circ}$. Thus, $B \widehat{A} D=\alpha=47^{\circ}$.

REMINDER 1: The base angles of an isosceles triangle are equal:

$\widehat{B}=\widehat{C} \quad$ and $A B=A C$.
REMINDER 2: The sum of the internal angles of a triangle is $180^{\circ}$ :

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-098.jpg?height=187&width=246&top_left_y=1734&top_left_x=1616)
$\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}$.