It is clear that the lengths of concurrent cevians cannot always determine the triangle. Indeed, they probably never can. But if it is clear, we must be able to give an explicit example.
Cevians $1$: Draw an equilateral triangle with height $1$. Pick as your cevians the altitudes.
Cevians $2$: Draw an isosceles triangle $ABC$ such that $AB=AC$, and $BC=\dfrac{10}{12}$, and the height of the triangle with respect to $A$ is equal to $1$. Then $AB=AC=\sqrt{1+(5/12)^2}=\dfrac{13}{12}$.
There are (unique) points $X$ and $Y$ on $AB$ and $AC$ respectively such that $BY=CX=1$. This is because as a point $T$ travels from $B$ to $A$ along $BA$, the length of $CT$ increases steadily from $\dfrac{10}{12}$ to $\dfrac{13}{12}$, so must be equal to $1$ somewhere between $B$ and $A$. Let $X$ be this value of $T$.
Let one of our cevians be the altitude from $A$, and let $BY$ and $CX$ be the other two cevians. These three cevians are concurrent because of the symmetry about the altitude from $A$, and they all have length $1$.
Reading farther on the Linked Wikipedia page, we have two cases:
Case 1: the triangle has an angle $\ge120^{\circ}$: the Fermat point is the obtuse-angled vertex.
Case 2: the triangle does not have an angle $\ge120^{\circ}$: the Fermat point is the first isogonic center. Following the link to this page, we find that point has trilinear coordinates $\csc(A+120^{\circ}):\csc(B+120^{\circ}):\csc(C+120^{\circ})$. If, like me, you are not familiar with trilinear coordinates, follow the link to its wikipedia page, where we find a method to convert trilinear coordinates to cartesian:
Taking vertex $C$ as the origin, find vectors $\vec{A}$ and $\vec{B}$ corresponding to the vertexes $A$ and $B$ respectively. Given side lengths $a, b, c$ corresponding in the usual way to the sides of the triangle respectively opposite vertices $A, B, C$, a trilinear coordinate $x:y:z$ is converted to a vector in cartesian coordinates by $$\vec{P}=\frac{ax}{ax+by+cz}\vec{A}+\frac{by}{ax+by+cz}\vec{B}.$$
In short, you have a piecewise function that checks for angle size and, if greater than 120 degrees, returns the $\vec{P}$, plus a correction $\vec{C}$ representing the location of $C$ if it is not convenient to have $C$ at the origin. You could embed the computation for $\vec{A}$ and $\vec{B}$ into the formula to have a single function result. That computation is necessary only if you don't already have coordinates for the vertices, and is elementary trig.
Best Answer
Denote the points like on this picture:
Consider the following composition of rotations: $I= R_{C',60^\circ}\circ R_{A',60^\circ}\circ R_{B',60^\circ}$. The classification of isometries says that $I$ is a central reflection, but also $I(A)=A$, so $I$ is the central reflection with respect to $A$: $I=S_A$. Thus $A$ can be constructed (e.g. take arbitrary $X$ and construct $X':=I(X)$; $A$ is the midpoint of $XX'$). In a similar way, by considering suitable compositions of rotations, you can construct $B$ and $C$.