This is a problem based on "Symmetry" of the plane $\mathbb{R^2}$. Suppose $A$, $B$, $C$ are the three points in plane which are after the corresponding actions by $s$ and $t$ are in the places $D$, $E$ and $F$ respectively. Let $G$ be any other point in the plane and and suppose $s$ and $t$ take it to $H$ and $I$ respectively. I need to find a contradiction somewhere.
So I join the sides and make some triangles with the hope of finding something. I proceed with the assumption that the original distance between any two points remains the same after the transformation. After this , Since $B$ is mapped to $E$ and $G$ is taken to $H$ and $I$, we have $EH=EI$. Similarly we have $DH=DI$ and $HF=FI$. So we have $\sphericalangle FHI=\sphericalangle FIH$, $\sphericalangle EHI=\sphericalangle EIH$ and $\sphericalangle DHI=\sphericalangle DIH$. Now $\sphericalangle I=\sphericalangle FHD \implies \sphericalangle FIE + \sphericalangle EID= \sphericalangle FHD \implies \sphericalangle EHD + \sphericalangle EID=0$
Thus $\sphericalangle EHD= \sphericalangle EID=0$ and we are done.
But this is a purely geometrical argument provided this is a correct one. I was wondering if there is an algebraic one.
Thanks for the help!!
I'll assume that since you assumed invariant distance in your geometric proof, by symmetries of the plane you mean isometries of the plane, i.e. elements of the Euclidean group in two dimensions. These can be represented by augmented matrices:
$$ \pmatrix{x\\y\\1}\to\pmatrix{a_{xx}&a_{xy}&t_x\\a_{yx}&a_{yy}&t_y\\0&0&1}\pmatrix{x\\y\\1}\;. $$
If two such isometries coincide for a point $(x,y)$, subtracting the two transformed points yields two homogeneous linear equations for the difference of the two augmented matrices:
$$ a_{xx}x+a_{xy}y+t_x=0\;,\\ a_{yx}x+a_{yy}y+t_y=0\;. $$
If there are three such points, the determinant of the resulting $6\times6$ system of linear equations is
$$ \left|\matrix{ x_1&y_1&1&0&0&0\\ x_2&y_2&1&0&0&0\\ x_3&y_3&1&0&0&0\\ 0&0&0&x_1&y_1&1\\ 0&0&0&x_2&y_2&1\\ 0&0&0&x_3&y_3&1\\ }\right|\;. $$
This determinant is zero iff the three points are collinear, and it follows that the difference of the two augmented matrices is zero if they aren't.