A: Yes! Joe was right. Start with a 45-45-90 triangle. Rotate by 90° about the right angle of the triangle. Then reflect across the line where the hypotenuse of the triangle used to be. The result is an isometry with no fixed points that is not a translation.

(I figured this out fairly quickly once I stopped trying to disprove it.)

**Corollary 1.4.4:** Suppose line l intersects two other lines
l_{1} and l_{2} so that the opposite interior angles
are equal. Then l_{1} and l_{2} are parallel.

**Discussion of proof:** We know that this result is true if the
angles in question are right angles. In order to prove it for other
angles, we construct some congruent right triangles, prove that they
are congruent, and then use the fact that they're right triangles to
apply Theorem 1.4.3.

**Proof:** Name the two points of intersection P and Q, with point
P on l_{1} and point Q on l_{2}. Construct segment PQ
and let O be the midpoint of that segment.

Choose a point R on line l_{1} so that line OR is
perpendicular to line l_{1} (Lemma 1.4.2 allows us to do this.
Ask if you want to see the proof.)

Since O is the midpoint of PQ, there is a 180° rotation about O
which maps P to Q. Let R' be the image of R under this rotation.
Note that we do not yet know whether R' is a point on line
l_{2}.

Look for congruent triangles: by choice of O, |OP| = |OQ|. Because isometries (rotations) preserve distance, |OR| = |OR'|. And because the rotational isometry takes one to the other, angle POR is congruent to angle QOR'. Therefore triangle POR is congruent to triangle QOR'.

By congruent triangles we know that angle OPR is congruent to angle
OQR'. Also, we started out with the condition that opposite interior
angles are equal -- angle OPR is congruent to the angle at the
intersection of PQ and l_{2}. This lets us conclude that that
angle of intersection of l and l_{2} and angle OQR' are
congruent, and so R' must lie on line l_{2}.

We chose R so that angle ORP would be a right angle. By congruent
triangles, angle OR'Q is also a right angle, so line RR' is
perpendicular to line l_{2}.

Since RR' is perpendicular to both lines l_{1} and
l_{2}, by Theorem 1.4.3 they must be parallel, which is what
we were trying to prove.//

**Corollary 1.4.5 (Euclid's Fifth postulate):** Suppose a line
meets two other lines so that the sum of the angles on one side is less
than two right angles. Then the other two lines meet at a point on
that side.

**Discussion of Proof:** We use Corollary 1.4.4 and our Axiom 5 --
given a point and a line, there is only one line through the point
that is parallel to the given line -- to conclude that the two lines
can't be parallel and so must intersect.

As you read the proof, notice how most of the work of the proof goes into constructing the lines and angles we need. Once the stage is set, we apply two theorems and complete the proof. This is typical of corollaries (as opposed to theorems).

**Proof:** Using notation similar to that of Corollary 1.4.4, let P
and Q be the points of intersection of l_{1} and l_{2}
with l. Let alpha be the angle of intersection between l and
l_{1}.

Consider the transformation along line l that takes P to Q (using our
standard assumptions about circles intersecting with lines, we can
prove this exists using axioms 6 and 7.) This transformation takes
line l_{1} to some line l_{3} through Q. Because we
used an isometry, the angle between l and l_{3} is the same as
that between l and l_{1}.

Because the measures of supplementary angles sum to 180° we can
show that we have equal alternate interior angles. By Corollary
1.4.4, l_{3} must be parallel to l_{1}.

Axiom 5 says that there can be only one line parallel to l_{1}
through point Q, and we know that lines l_{2} and
l_{3} are different because they must meet line l at different
angles. (Remember, "the sum of the angles on one side is less than
two right angles".)

Since l_{3} is parallel to l_{1}, l_{2} must
not be. So l_{2} must meet l_{1} at some point R.

(We've now proved all of the theorem except the part that says "on that side".)

On one side of l the angles sum to more than 180°. On this side,
line l_{3} lies between lines l_{1} and l_{2}.
If line l_{2} meets line l_{1} on this side,
l_{3} must either cross segment PR or recross line
l_{2}. (By some complicated stuff that is proven in Chapter
9) neither of these is possible, so point R must lie on the side of l
on which the angles sum to less than 180°.//

We've proven that if we assume Axiom 5 we can prove Euclid's fifth postulate! To prove that it doesn't matter whether we start by assuming Axiom 5 or the fifth postulate, we'd still have to start with the fifth postulate and prove Axiom 5. That's Exercise 1.21.

Theorem 1.4.1:Let P be a point not on l, and let Q lie on l so that PQ is perpendicular to l. Let l_{2}be the line through P which is parallel to l (as guaranteed by our Axiom 5). Then l_{2}intersects PQ at 90°.

**Discussion of the Proof:** Start by drawing a (possibly
misleading) sketch of all the lines and points mentioned in the
theorem statement. This will help you follow the proof.

The proof proceeds **by contradiction**. In other words, the
author starts out by saying "what if line l_{2} *isn't*
perpendicular to line PQ?" He shows that if the lines aren't
perpendicular there must be *another* line through P
that is parallel to l, which violates our basic assumption (Axiom 5)
that the line through P parallel to l is unique. Since Axiom 5 is
assumed to be true, an assumption that proves it false must have been
false to begin with. To summarize: if l_{2} is not perpendicular
to PQ then Axiom 5 is false; we're assuming Axiom 5 is true, so the two
lines must be perpendicular.

We'll need the following lemma to support our proof:

**Lemma A:** If a line l_{1} ≠ l is sent to itself under a
reflection through l, then l_{1} and l intersect at right
angles.

**Proof of Lemma A:** Consider one of the angles formed by the
intersection of l_{1} and l. The image of that angle under
reflection through l is a congruent angle, because reflection is an
isometry. Definition 4 on page 19 shows that those angles must be
right angles.//

Now we're ready to go over the book's proof that a line perpendicular to a line l must be perpendicular to all lines parallel to l. Start by drawing a (possibly misleading) sketch of all the lines and points mentioned in the theorem statement. This will help you follow the proof.

**Proof:** Suppose that l_{2} does not intersect line PQ at an angle of
90°. Then (using Lemma A), when you reflect l_{2} through
line PQ you get a new line l_{3}. Note that when you reflect l
through line PQ you do get line l (by Definition 4.)

Since we chose l_{2} so that it does not intersect line l,
line l_{3} must not intersect l either. But that means we
have *two different* lines through P which do not intersect l,
which contradicts our Axiom 5.

Since we reach this contradictory conclusion whenever the angle
between l_{2} and PQ is not 90°, it follows that that
angle must always be 90°.//

Any questions?

I think we cheated when we picked our point Q on line l. We don't
automatically know that we can draw a perpendicular line through P to
l. However, the book clears this up by proving Lemma 1.4.2 after the
fact. (It's unusual to write a book in which a lemma appears
*after* the theorem is written, but it's common practice to
prove lemmas after the theorem is mostly proved. A proof is just a
line of reasoning saying "if A is true and B is true then this C be
true; therefore D is true". It's very common to come up with a line
of reasoning like this, then go back and prove lemmas to show A and B
are true.)