Let $ABC$ be an isosceles triangle at $A$ and $D$ the foot of the internal bisector from $B$. Suppose that $BC = AD + DB$. What is the value of $\widehat{BAC}$?
(resolved independently by Antoine Bourion and Silvia Maris)

Let $E$ be the point of intersection of the internal bisector from $C$ which intersects the segment $[A B]$ at a point $E$. We place $F$ on $(B C)$ such that $B D = B F$, and we call $\alpha$ the angle $\widehat{C B D}$.

The line $(D E)$ is parallel to $(B C)$ since $B$ and $C$ are symmetric with respect to the altitude from $A$ in $A B C$.

Thus, $\widehat{D E A} = \widehat{E D A} = \widehat{C B A} = 2 \alpha$. We set $\beta = \widehat{B A C}$, and we have $\beta = 180 - 4 \alpha$.

The equality $B C = A D + D B$ implies $A D = A E = F C$. Furthermore, the triangle $C D E$ is isosceles at $D$ since $\widehat{D C E} = \widehat{D E C} = \alpha$ by alternate interior angles. The triangles $E D A$ and $D C F$ are similar because the angles $A D E$ and $F C D$ are equal, so the triangles have two lengths and one angle in common. Moreover, $A D = F C$ and $D F = C D$.

The triangle $C D F$ is therefore isosceles at $F$, the angle $\widehat{C D F}$ is $2 \alpha$ and the angle $\widehat{C F D} \beta$.

From this, $\widehat{B D F} = \widehat{B F D} = 180 - \beta = 4 \alpha$ and thus,

$$
180 = \widehat{A D E} = \widehat{E D B} = \widehat{B D F} = \widehat{F D C} = (2 + 1 + 4 + 2) \alpha
$$

We deduce that $4 \alpha = 90$ and $\beta = 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_00138f973c2e12c014a0g-432.jpg?height=971&width=1676&top_left_y=857&top_left_x=201)