In the figure, the triangle $\triangle A B C$ is isosceles, with $B \widehat{A} C=20^{\circ}$.

Knowing that $B C=B D=B E$, determine the measure of the angle $B \widehat{D} E$.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-006.jpg?height=463&width=211&top_left_y=254&top_left_x=1619)
By definition, a triangle is isosceles if it has two equal sides. The third side is called the base of the isosceles triangle, and the angles formed between the base and the two equal sides are the base angles.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-041.jpg?height=448&width=443&top_left_y=267&top_left_x=818)

The figure shows an isosceles triangle $\triangle ABC$, whose equal sides are $AB$ and $AC$ and the base is $BC$. We denote the base angles $A \widehat{B} C$ and $A \widehat{C} B$ by $\widehat{B}$ and $\widehat{C}$, respectively. It is proven that in an isosceles triangle, the base angles are always equal. In the triangle in the figure, we therefore have $\widehat{B} = \widehat{C}$.

Moving on to solving this problem, observe that $\widehat{A} + \widehat{B} + \widehat{C} = 180^{\circ}$, since the sum of the interior angles of any triangle is $180^{\circ}$. Given that $\widehat{A} = 20^{\circ}$ and the triangle is isosceles, so $\widehat{B} = \widehat{C}$. Therefore, $180^{\circ} = 20^{\circ} + \widehat{B} + \widehat{C} = 20^{\circ} + 2 \widehat{B}$, and thus $\widehat{B} = \widehat{C} = 80^{\circ}$.

The triangle $\triangle CBD$ is also isosceles, as it is given that $CB = DB$. Since the base of this triangle is $CD$, its base angles are $C \widehat{D} B = \widehat{C}$, so $C \widehat{D} B = 80^{\circ}$. Considering the sum of the interior angles of this triangle $\triangle CBD$, we get $C \widehat{B} D + C \widehat{D} B + \widehat{C} = 180^{\circ}$. Substituting the values already obtained, we see that $C \widehat{B} D + 80^{\circ} + 80^{\circ} = 180^{\circ}$, so $C \widehat{B} D = 20^{\circ}$. Thus, $D \widehat{B E} = \widehat{B} - 20^{\circ} = 80^{\circ} - 20^{\circ} = 60^{\circ}$.

The triangle $\triangle DBE$ is also isosceles, because $DB = BE$. The base of this triangle is $DE$ and the equal base angles $BE$ are $E \widehat{D} B = D \widehat{E} B$. Since

$$
180^{\circ} = B \widehat{D} E + D \widehat{E} B + D \widehat{B} E = 2 \times B \widehat{D} E + 60^{\circ}
$$

we conclude that $B \widehat{D} E = 60^{\circ}$.