I have some problems in the following task of my geometry studies:
Let $A,B,C$ three non-collinear points in $\mathbb{R}^2$ and let $T_1,T_2:\mathbb{R}^2\rightarrow\mathbb{R}^2$ two congruence transformations so, that:
- $T_1(A)=T_2(A)$
- $T_1(B)=T_2(B)$
- $T_1(C)=T_2(C)$
Show that $T_1=T_2$.
I tried using the definition of congruence transformations and that the distances between these image points don't change under $T_1$ or $T_2$. But i don't know how to go further. Can anybody help me ?
Lemma Let $T$ be congruence transformation in a plane and let for every $X$ be $X' :=T(X)$. Then for any three points $K,L,M$ are collinear iff $K',L',M'$ are collinear.
Proof: Say $K,L,M$ are collinear and suppose $K',L', M'$ are not. Then $$K'L' + L'M' > K'M'$$ where $K'M'$ is the largest among $K'L', L'M' $ and $K'M'$. But then $$KL + LM > KM$$ A contradiction. The same proof works for the other way.
Let for an arbitrary point $X$ in the plane $X'=T_1(X)$ and $X'' = T_2(X)$. So we have $A'=A''$, $B'=B''$ and $C'=C''$.
Now we have to prove that for every $X$ in the plane we have $X'=X''$. If $X'\ne X''$ then, since $$A'X' = A''X'' = A'X''$$ we se that $A'$ lies on perpendicular bisector for $X'X''$. But the same is true for $B'$ and $C'$. So $A',B', C'$ are collinear and so are $A,B,C$. A contradiction.