Theorem in book states that if (A,B,C) and (A', B', C') are two ordered sets of three non-colinear points in plane alpha and if is AB=A'B', BC = B'C' and AC=A'C', then there is isometry I which translates plane alpha into plane alpha: I(alpha) = alpha and then is I(A) = A', I(B) = B' and I(C) = C'.

I don't have proof of that theorem in book so I tried to prove it.

My proof is based on proving that for any given line segment in plane alpha in regard to A,B,C there is congruent line segment in plane alpha in regard to A',B',C'. There can be more cases of where line segment can be located but I will consider just one.

1) If there is some line segment MN in plane alpha then line that contains MN must intersect at least two lines formed by A,B,C points. Lets say that line MN intersects line BC and line AC in points X and Y. Then we can find X' and Y' on lines B'C' and A'C' such that (A,C,Y) = (A',C',Y') and (B,C,X) = (B',C',X'). Then AC=A'C', CX = C'X' and AX = A'X'.

Since AY = A'Y' we have that from (A,C,X) = (A',C',X') follows that XY = X'Y'. From that we can easy find and prove that MN=M'N'.

From that follows that I(alpha) = alpha.

Can someone look at my proof?