What we'll prove is that if the two sides of a triangle are the same length, the angles those sides make with the third side must also be equal. Euclid proves slightly more than this, because he can.

Euclid first introduces and names some new points to help in the proof. Points D and E are just points "further down the line" and don't figure in the rest of the proof. Point F can be any point on line AD, and G is a point on line AE chosen so that |AF| = |AG|. (How do we know there is such a point? Postulate 3 allows Euclid to define a circle with radius |AF| and center A. Euclid then assumes (without justification) that that circle will intersect line AE at some point, which he names G.)

He then uses Proposition 4, which he has already proven and which we know as "Side Angle Side", to show that triangles AFC and AGB are congruent. Because of this, he can conclude that |GB| = |FC| and that the angles at F and G are congruent, and so he can again use Proposotion 4 and conclude that triangles BGC and CFB are congruent.

Since m∠FCA = m∠GBA and m∠FCB = m∠GBC, Euclid concludes that m∠ABC = m∠ACB.//

The author of our textbook prefers to use a different set of tricks
from the ones Euclid used. His preferred tools are isometries;
**isometries are flips or sliding motions that superimpose one
object on another object of equal proportions.** On page 13 the
author gives a more precise definition of isometries as motions of the
plane which preserve all distances between points in the plane.

The first proof we will look closely at says that if the side lengths of one triangle match the side lengths of another, those two triangles are congruent ("Side side side"). Since we're trying to take as little for granted as possible, we want to define the word "congruent" if we can.

In fact, we can use the idea of an isometry to define congruence! If
we have one figure, and we can flip and slide it around until it
exactly coincides with another figure, we say that those two figures
are congruent. **Two sets of points are congruent if there exists an
isometry that maps one to the other.**

I've talked a lot about "flipping and sliding", which takes many things for granted. The book is more careful with its fancy definition involving "distance". Axioms 6 through 8 on page 13 of the book describe the assumptions the author needs to make about isometries in order to use them the way he wants to. Axioms 6 through 8 also happen to describe different ways of flipping and sliding figures around in the plane.

Axiom 6: Given any points P and Q, there exists an isometry f so that
f(P) = Q.

This just says that there's some way to slide (translate), turn (rotate) or flip (reflect) any point in the plane onto any other point.

Axiom 7: Given a point P and any two points Q and R which are
equidistant from P, there exists an isometry which fixes P and sends Q
to R.

Suppose you have three points arranged as shown:

A B Cwith |AB| = |AC|. Then you can either reflect (flip) across the vertical line through A to put B on to C, or you can rotate (spin) point B around point A until it lands on top of C. (This rotation would move C to a point above and to the right of where it is now -- the fact that isometries preserve distances between points means that C has to end up as far from B as it was before we moved B.)

In other words, you pick a point to stay still and a point to move around it to another point. There is a way to move the points like this (an isometry) only when the point you're moving to is the same distance away from the fixed (still) point as the point you started with.

Axiom 8: Given any line l, there exists an isometry which fixes
every point in l but fixes no other points in the plane.

Reflecting (flipping) across a line is an isometry.

There are formal definitions of reflection, translation and rotation on page 16. (I believe that the author chooses isometries over the traditional tools from ruler and compass constructions because he prefers a transformational approach to teaching geometry. It's possible to teach high school geometry using either a classical or a transformational approach, depending on the materials available.)

Homework: Read page 15 and post a question about it on the message board.

How are we going to prove this? The definition of "congruent" tells us we have to find an isometry that maps one triangle to the other. It's too hard to do this all at once, so we do it step by step, finding an isometry for each step. (You'll convince yourself for homework that if you do one isometry then another, the resulting combination is also an isometry.)Theorem 1.3.1:If the corresponding sides of two triangles ΔABC and ΔA'B'C' have equal lengths, then the triangles are congruent.

**Proof:** It turns out that the proof the author uses doesn't work
if the triangle is degenerate (all vertices are on the same line). He
probably realized this after he finished proving it, but when he wrote
down the final version of his proof he mentioned it at the beginning.
The proof is left to the reader. In this case it's a fairly easy
proof, but if you ask me in advance you might get a tiny bit of extra
credit for writing it up and turning it in.

Here I'll go through the rest of the proof step by step, trying to
relate make each isometry a separate step. Our plan is to prove that
we can slide, flip and turn the plane until vertices A, B and C land
on top of vertices A', B' and C', respectively.
Step 1: Use isometry f_{1} to move A to A'.

Axiom 6 says there exists an isometry that does this. Let's call it
f_{1}. (Notice that all we need to know is that it exists; we
don't have to know exactly what it is. This is common in mathematical
proofs, and can lead to the odd situation of knowing something is
possible but not how to do it.)

Step 2: Hold f_{1}(A) at A' and move f_{1}(B) to
B'.

Here we use Axiom 7. We have a point A' = f_{1}(A). We also
have two points f_{1}(B) and B' that are equidistant from A'.
(We know these distances are equal because we started out knowing that
|AB| = |A'B'|, and isometries don't change distances.) Axiom 7 says
there exists an isometry that fixes (doesn't move) A' and sends
f_{1}(B) to B'. The author calls this isometry
f_{2}.

So f_{2}(f_{1}(A)) = A' and
f_{2}(f_{1}(B)) = B'. If
f_{2}(f_{1}(C)) = C' we're done. Otherwise, we need to:

Step 3: Move f_{2}(f_{1}(C)) to C' without changing
anything on line A'B'.
This step looks pretty easy. We have to leave
f_{2}(f_{1}(A)) = A' and
f_{2}(f_{1}(B)) = B' where they are and move
f_{2}(f_{1}(A)) = A' and
f_{2}(f_{1}(C)) to C'. Reflecting the whole plane
through line A'B' ought to do that.

We know that such a reflection exists, and we know that that
reflection will take f_{2}(f_{1}(C)) to a point that's
distance |AC| from A' and to a point that's distance |BC| from B'.
That describes point C', so that ought to finish the proof.

**However** we don't know that there isn't some sneaky point D
that's also distance |AC| from A' and distance |BC| from B'. Lucky
for us, the author proved in Chapter 9 that if two circles intersect,
they intersect in at most two points.

We draw a circle around A' with radius |AC| and a circle around B'
with radius |BC|. These two circles must intersect at points C and
C'. Using the lemma the author proved in Chapter 9, we know the
circles don't intersect at any other points, so there are no other
points the right distances from A' and B'. Isometries preserve
distances, and we can use Axiom 8 to find an isometry f_{3}
that holds A' and B' in place while f_{2}(f_{1}(C))
moves, so that isometry must move f_{2}(f_{1}(C)) to
C'.
To summarize, we have proved that ΔABC and ΔA'B'C' are
congruent by showing that there is an isometry f_{3}
o f_{2} o f_{1} that
maps A to A', B to B' and C to C'.//

Homework: Explain why a combination of two isometries (e.g. a translation followed by a rotation) is still an isometry.