Find $B \widehat{A} D$, 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_dbce7bc2a200e234c3e2g-08.jpg?height=368&width=545&top_left_y=1838&top_left_x=1126)

## List 9
Given $A B=A C$, the triangle $\triangle A B C$ is isosceles, hence $A \widehat{B} C=A \widehat{C} B$. Since $A D=B D$, the triangle $\triangle A B D$ is also isosceles, hence $A \widehat{B} D=B \widehat{A} D$. Therefore,

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

![](https://cdn.mathpix.com/cropped/2024_05_01_dbce7bc2a200e234c3e2g-33.jpg?height=241&width=447&top_left_y=1099&top_left_x=1273)

In the figure, these 3 equal angles are represented by the letter $\alpha$. The internal angles of $\triangle A B C$ are $\alpha+39^{\circ}, \alpha$ and $\alpha$; thus:

$$
\alpha+39^{\circ}+\alpha+\alpha=180^{\circ} \Rightarrow 3 \alpha=180^{\circ}-39^{\circ}=141^{\circ}
$$

Therefore, $B \widehat{A} D=47^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_dbce7bc2a200e234c3e2g-33.jpg?height=218&width=662&top_left_y=1723&top_left_x=344)

REMINDER 2: In an
isosceles triangle, the
base angles are equal:
$\widehat{B}=\widehat{C}$ and $A B=A C$.

## List 9