[ Equilateral (equiangular) triangle ]

$[\quad$ Method GMT $]$

Given an equilateral triangle ABC. Find the set of all points $D$ such that triangles $ABD$ and $BCD$ are isosceles (segments AB and BC can serve as either the bases or the legs).
The set of all vertices of isosceles triangles with base $AB$ is the axis of symmetry of segment $AB$, and the set of vertices of isosceles triangles with base $BC$ is the axis of symmetry of segment $BC$. The intersection point of these two axes (see figure) gives the first desired point $D_{1}$. However, sides $AB$ and $BC$ can also be the legs of the desired isosceles triangles with a common vertex $D$. If the vertex of isosceles triangles $ABD$ is point $A$, then the set of points $D$ will be a circle with radius $AB$ centered at point $A$. The points of intersection of this circle with the axis of symmetry of segment $CV$ and the circle centered at point $C$ with radius $BC$ will give the points $D_{2}, D_{3}$, and $D_{4}$, respectively. Considering isosceles triangles $BCD$ with vertex $C$, we find the points of intersection of the circle centered at point $C$ with radius $CV$ and the axis of symmetry of segment $AB$, giving two more desired points $D_{5}$ and $D_{6}$. Finally, the set of points $D$ that are common vertices of isosceles triangles $ABD$ and $CBD$ with vertex $B$ is a circle centered at point $B$ with radius $BC$ (or, equivalently, radius $BA$). Thus, the desired set of points consists of the specified circle and six individual points. The points on the circle exclude the vertices of the triangle $A$ and $C$ and the points of intersection with the extensions of sides $AB (F)$ and $BC (E)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-09.jpg?height=731&width=806&top_left_y=838&top_left_x=0)