Congruence
Axiom 1: If A and B are distinct
points and if A' is any point, then for each ray *r* emanating
from A' there is a unique point B' on *r* such that B' is
not equal to A' and AB is congruent A'B'.

Congruence Axiom 2: If AB is congruent to CD and AB is congruent to EF then CD is congruent to EF.

Congruence Axiom 3: If A*B*C, A'*C'*C', AB congruent to A'B', and BC congruent to B'C', then AC is congruent to A'C'.

Congruence Axiom 4: Given any angle BAC (where by definition of "angle," ray AB is not opposite to ray AC), and given any ray A'B' emanating from a point A', then there is a unique ray A'C' on a given side of line A'B' such that angle B'A'C' is congruent to angle BAC.

Congruence Axiom 5: If angle A congruent to angle B and angle A congruent to angle C, then angle B congruent to angle C.

Congruence Axiom 6: If two sides and the included angle of one triangle are congruent respectively to two sids and the incuded angle of another triangle, then the two triangles are congruent.

AB<CD, AB congruent to CD, or AB > CD.

(1) AB < CD

(2) AB congruent to CD

(3) AB > CD

We know that B' ? C , so by definition of ray at least one of the following is

true:

(a) C * B' * D

(a) and (c) cannot both be true by betweeness axiom 1

(b) and (c) cannot both be true by betweeness axiom 1.

Therefore, exactly one of (a), (b), and (c) are true.

(a) if and only if (1)

(b) if and only if (2)

(c) if and only if (3)

(a) if and only (1):

(b) if and only if (2):

To see if (b), then (2), assume D = B'. Then CB' congruent to CD, by congruence axiom 2 and AB congruent to CB'. So, AB congruent to CB' by congruence axiom 2.

AC congruent to DF, for any point B between A and C there is a unique point E between D and F, such that AB congruent DE, there is a unique point D' between A and B so that CD congruent to AD'. So, by definition of <, CD < AB.

Now, to see if (3), then (c), assume CD < AB. Then there exists D' between A and B so that AD' congruent to CD. We know AB congruent to CB', so also by proposition 3.12, there is a point D'' so that C*D''*B and AD' congruent to CD''. Then by congruence axiom 2, CD congruent to CD''. So by congruence axiom 1, D = D'' and therefore C*D*B'.

AB < CD, AB congruent to CD, or AB > CD.

Given AB < CD, they by definition of <, there exists a point G on CD such that C*G*D and AB @ CG. By proposition 3.12, there exists a point G' on EF, such that CG congruent to EG'. Then by congruence axiom 2, AB congruent to EG'.

Therefore, AB < EF by definition.

CD < AB. Then by definition, there exists a point G on AB such that A*G*B and AG congruent to CD. Since CD congruent to EF and CD congruent to AG then by congruence axiom 2,

EF congruent to AG.

Definition: Angle ABC < Angle DEF means there is a ray EG between ray BA and ray BC, ray EH between ray ED and ray EF, angle CBG congruent to Angle FEH and Angle ABC congruent to Angle DEF. Then Angle GBA is congruent to Angle HED.

Link to geometers sketchpad so you can draw these propositions