There are infinite many equilateral pentagons IJKLM, inscribed in a triangle ABC, as shown.

Essentially all these pentagons are parameterized by the angle (psi), determining the slope of side IJ to AB. Taking D arbitrary on AB and drawing DE parallel to NO (realizing the slope (psi)), we find E. Then from D and E we lay correspondingly segments equal to DE, finding points H and F. Then construct the isosceles with basis on HF and sides HG=GF=DE. This completes the construction of an equilateral pentagon DEFGH. The only shortcoming of the pentagon is that G is not on BC, in general. The clue is that all pentagons constructed that way, starting with an arbitrary point D on AB, are similar to each other and their vertices G lie on a fixed line [AL]. This determines point L on side BC, which, in turn, determines the pentagon IJKLM. Later e.g. by taking the similarity with center at A and ratio k = ML/HG and applying it to the pentagon DEFGH. The existence of G depends on the validity of HF < 2DE.

Info on animation: Points A, B, C are free movable (switch to the selection tool: ctrl+1). Moving them changes the shape of the triangle. Points N, O and D are modifiable through the tool [Select on contour] (ctrl+2). Moving N, O changes the slope (psi). Moving D changes DEFGH by similarity.
The subject handled here is a particular one of a more general problem, initiated at the DivisionProblem.html .

For the determination of all the equilateral pentagons inscribed in a triangle one can start by asking how are distributed the vertices on the sides of the triangle. The present case is the main one, for which the distribution is (2,2,1). There is also the case (1,1,3), which is degenerate. The pentagon has three of its vertices on one side of the triangle and degenerates to a quadrangle inscribed in the triangle and having three sides equal to t, say, and the other equal to 2t. This is handled in PentadivisionDegenerate.html .

Look at the file Hexadivision.html for an analogous discussion for inscribed hexagons.