Does congruence guarantee length conversion? Suppose that a linear transformation $M:R^2 \rightarrow R^2$ maps a triangle $ABC$ to a congruent triangle $A'B'C'$
($\{A, B, O\}, \{B, C, O\},\{C, A, O\}$ are not colinear, and $A,B,C\neq O$)
Is it true that the linear transformation always conserves length?
Or can there possibly be a counterexample?
(I'm talking about transformations on $R^2$ that can be represented as a 2 by 2 matrix)
Thanks in advance
 A: It is true, whether or not "linear" means homogeneous-linear taking $0$ to $0$.
Triangle $ABC$, if it is nondegenerate (nonzero area), determines a linear coordinate system in which $A=(0,0)$, $B=(0,1)$ and $C=(1,0)$, and coordinate $(x,y)$ is assigned to the point $A + x(B-A) + y(C-A)$.  The distances between pairs of points are fully and uniquely determined by the coordinates of the points and the lengths $AB, BC$ and $CA$.  Therefore, the unique invertible affine (linear inhomogeneous) transformation

"for all $(x,y)$, send the point with $ABC$-coordinates $(x,y)$ to the point with $A'B'C'$-coordinates $(x,y)$"

is an isometry if $ABC$ is congruent to $A'B'C'$ and both triangles are non-degenerate.  It is also the unique affine-linear transformation sending $ABC$ to $A'B'C'$, so that any linear transformation with that property is an isometry.
A: By definition, all congruent triangles have the same size. So you can map the sides $AB$, $BC$, $CA$ to $A'B'$, $B'C'$ and $C'A'$, respectively. Now let's consider if this mapping is an isometry. Clearly, $|AB|=|A'B'|$, as we have defined by congruency. So the vector $AB$ is mapped to another vector of the same length. The same happens to the vector $BC$ mapped to $B'C'$. These vectors however are linearly independend, as you defined. Now you use the fact that a linear mapping is a isometry iff the mapping of each basis vector keeps the original length constant.
