Let $A,B,C$ be three non-collinear points. I want to prove this without the result that if a isometry fix three points then it is identity.
I thought on doing this: Suppose $f(D) \neq g(D)$. Then, as f and g are isometries, $AD \equiv Af(D) \equiv Ag(D), BD \equiv Bf(D) \equiv Bg(D), CD \equiv Cf(D) \equiv Cg(D)$. I can "see" that this situation cannot happen in the plane, but I'm having thouble writting this absurd. Can someone please show me how to conclude? Thanks
Denote $\overline{xy}$ the line containing points $x$ and $y$.
Since the distances from $f(D)$ and from $g(D)$ to the points $f(A)=g(A)$ and $f(B)=g(B)$ are the same, $g(D)$ lies in the line $L_1$ which is perpendicular to $\overline{f(A)f(B)}$ and contains $f(D)$. Similarly, with $C$ in place of $B$, $g(D)$ lies in the line $L_2$ which is perpendicular to $\overline{f(A)f(C)}$ and contains $f(D)$.
But $f(A),f(B),f(C)$ are noncolinear, so $L_1$ and $L_2$ are distinct lines which intersect at $f(D)$, so this in in fact the only intersection point. Since $g(D)$ is in their intersection, $g(D)=f(D)$.