SSS for normed plane

Question

It is known that if in $(R^2,\|\cdot\|_2)$ is given two triangles with vertices $A,B,C$ and $D,E,F$ such that $\|A-B\|_2=\|D-E\|_2$, $\|B-C\|_2=\|E-F\|_2$ and $\|A-C\|_2=\|D-F\|_2$ then by Heron's formula it follows that $S_{ABC}=S_{DEF}$, where $S_{XYZ}$ area of triangle $XYZ$ and $\|\cdot\|_2$ usual euclid norm. What would be if we instead of $\|\cdot\|_2$ consider any norm $\|\cdot\|$? In other words, let $A,B,C$ and $D,E,F$ respectively vertices of two triangles in $(R^2,\|\cdot\|)$, such that $\|A-B\|=\|D-E\|$, $\|B-C\|=\|E-F\|$ and $\|A-C\|=\|D-F\|$. Is it true that $S_{ABC}=S_{DEF}$, where $S_{XYZ}$ is "usual" area of the triangle with vertices $X,Y,Z$?

Answer 1

It seems to be satisfied, if $\|\cdot\|$ is generated by an inner product. To give an example, let $T \in \mathbb R^{2 \times 2}$ be invertible. We use the norm $$\|x\|^2 = x^\top T^\top T \, x.$$

Then, if your triangles $(A, B, C)$ and $(D, E, F)$ satisfy your condition w.r.t. $\|\cdot\|$, it is easy to see that $(TA, TB, TC)$ and $(TD, TE, TF)$ satisfy the condition w.r.t. $\|\cdot\|_2$. Hence, the transformed triangles have equal areas, since the transformation $T$ scales area by $\det(T)$.

In other norms, it might be violated. For example, you can use $$A = D = (0,0), B = (1/2, 1/2), C = (0,1), E = (1/4,3/4), F = (3/4, 1/4),$$ with the $1$-norm. All sides have length one, one triangle has area $1/4$ and the other one has $1/2$.