$:$ Folklore

In triangle $A B C$, angle $B$ is equal to $60^{\circ}$. Point $D$ inside the triangle is such that $\angle A D B=\angle A D C=\angle B D C$. Find the smallest value of the area of triangle $A B C$, if $B D=a$.
From the condition, it follows that $\angle A D B=\angle A D C=\angle B D C=120^{\circ}$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_476f7e28c93a861c2173g-30.jpg?height=515&width=644&top_left_y=0&top_left_x=707)

From triangle $A B D: \angle D A B+\angle D B A=\angle B=60^{\circ}$, hence, $\angle D A B=\angle D B C$. Therefore, triangles DAB and $D B C$ are similar. Thus,

$A D: B D=B D: C D$, which means $A D \cdot C D=B D^{2}=a^{2}$. Therefore, $S_{A D C}=A D \cdot C D \cdot \sin 120^{\circ}$ is a constant value.

Thus, the area of triangle $A B C$ will be the smallest if the sum of the areas of triangles $A D B$ and $B D C$ is the smallest. But

$S_{A D B}+S_{B D C}=1 / 2 a(A D+C D) \sin 120^{\circ}$, and the sum $A D+C D$ is minimal when triangle $A D C$ is isosceles (see problem $\underline{32883}$ b).

In this case, obviously, triangle $A B C$ is equilateral and $S_{A D B}=S_{B D C}=S_{A D C}=\frac{a^{2} \sqrt{3}}{4}$.

## Answer

$\frac{3 a^{2} \sqrt{3}}{4}$.