1. ## Proof of isometry of plane (please verify)

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?

2. Just to actualize my thread which hasn't get reply for 5 days since posting.

Can someone look at it?

3. I cannot verify what you have done because I am not sure of what axioms you are following. However, it seems that you may have access to a reasonably good mathematics library. If so find Modern Geometries by James R. Smart. In that text there is a very good proof of the following theorem: A motion of the plane is uniquely determined by an isometry of one triangle onto another.

4. Originally Posted by Plato
I cannot verify what you have done because I am not sure of what axioms you are following.
I am following axioms of incidence, order, parallels and congruence.

Relevant axiom of congruence used in this proof is:
If $A,B,C$ and $A',B',C'$ are two sets of three non-colinear points and $D$ and $D'$ are points on rays $BC$ and $B'C'$ such that $(A,B) \cong (A',B')$, $(B,C) \cong (B',C')$, $(C,A) \cong (C',A')$ and $(B,D) \cong (B',D')$ then is also $(A,D) \cong (A',D')$.

And also theorem that if A,B,C are three points on line m and A',B' points on line m' such that AB=A'B' then there is unique point C' on line m' such that is AC=A'C' and BC=B'C'.

I am not sure of what axioms you are following. However, it seems that you may have access to a reasonably good mathematics library. If so find Modern Geometries by James R. Smart. In that text there is a very good proof of the following theorem: A motion of the plane is uniquely determined by an isometry of one triangle onto another.
I have only two books on Geometry. One is high-school book and one is university book on Euclidian and hyperbolic geometry.
I would like to have good math library on Geometry, but money is little problem... Only what I can find free on Internet.

5. Have you tried a local public library?
I am not sure in what part of the world you are?
But most large city libraries are pretty good.

6. OReilly I see that you are learning "Modern Geometry" (that is the way I heard it is called) it looks really interesting, hope to one day know it. But why did you chose it out of all the possibilities you had? It seems to me that this branch of math is one of the least studied (maybe set/logic theory are even more rare).

For the great Plato, where did you learn it from? College of by thyself from a textbook?

7. Originally Posted by ThePerfectHacker
OReilly I see that you are learning "Modern Geometry" (that is the way I heard it is called) it looks really interesting, hope to one day know it. But why did you chose it out of all the possibilities you had? It seems to me that this branch of math is one of the least studied (maybe set/logic theory are even more rare).

For the great Plato, where did you learn it from? College of by thyself from a textbook?
Can you explain me what exactly do you mean by "Modern Geometry" (that is the way I heard it is called)?

8. Originally Posted by OReilly
Can you explain me what exactly do you mean by "Modern Geometry" (that is the way I heard it is called)?
That is a more formal treatment of Euclidean Geomtery.

It may consist of non-Euclidean also, which is what you are doing.

9. For what 'plane' or surface are you asking if this works? Depending on the surface, you will have different conditions.

*EDIT*

Nevermind, I just realized this isn't Advanced Geometry and it's based on Euclidean.

10. Originally Posted by AfterShock
For what 'plane' or surface are you asking if this works? Depending on the surface, you will have different conditions.

*EDIT*

Nevermind, I just realized this isn't Advanced Geometry and it's based on Euclidean.

Yes it's based on Euclidean.

Just to actualize it again, since I haven't get answer from last time that I have actualize it.

12. Originally Posted by OReilly
That statement is amazing.