Let a triangle $ABC$ be given (figure below). Let us take an arbitrary point $P$, inside the triangle. Let us now join $P$ with the vertices $A$, $B$, $C$, so as to obtain segments $i=AP$, $j=BP$ and $k=CP$. I want to prove that if for any point $P$ it is possible to construct a triangle having its sides equal to $i$, $j$, and $k$, respectively, then the original triangle $ABC$ is equilateral.
To form a triangle, values $i,j,k$ have to satisfy equations: $$ i + j \geq k $$ $$ i + k \geq j $$ $$ j + k \geq i $$
If $P \rightarrow A$, then $i \rightarrow 0$, $j \rightarrow AB$ and $k \rightarrow AC$. The inequalities above then imply that $AB \geq AC$ and $AC \geq AB$, which means that $AB=AC$. In a similar way one can prove that $AB=BC$.