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 l1 and l2 so that the opposite interior angles are equal. Then l1 and l2 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.As always, make a sketch of each step as you follow the proof.
Proof: Name the two points of intersection P and Q, with point P on l1 and point Q on l2. Construct segment PQ and let O be the midpoint of that segment.
Choose a point R on line l1 so that line OR is perpendicular to line l1 (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 l2.
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 l2. This lets us conclude that that angle of intersection of l and l2 and angle OQR' are congruent, and so R' must lie on line l2.
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 l2.
Since RR' is perpendicular to both lines l1 and l2, 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 l1 and l2 with l. Let alpha be the angle of intersection between l and l1.
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 l1 to some line l3 through Q. Because we used an isometry, the angle between l and l3 is the same as that between l and l1.
Because the measures of supplementary angles sum to 180° we can show that we have equal alternate interior angles. By Corollary 1.4.4, l3 must be parallel to l1.
Axiom 5 says that there can be only one line parallel to l1 through point Q, and we know that lines l2 and l3 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 l3 is parallel to l1, l2 must not be. So l2 must meet l1 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 l3 lies between lines l1 and l2. If line l2 meets line l1 on this side, l3 must either cross segment PR or recross line l2. (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 l2 be the line through P which is parallel to l (as guaranteed by our Axiom 5). Then l2 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 l2 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 l2 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 l1 ≠ l is sent to itself under a reflection through l, then l1 and l intersect at right angles.
Proof of Lemma A: Consider one of the angles formed by the intersection of l1 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 l2 does not intersect line PQ at an angle of 90°. Then (using Lemma A), when you reflect l2 through line PQ you get a new line l3. Note that when you reflect l through line PQ you do get line l (by Definition 4.)
Since we chose l2 so that it does not intersect line l, line l3 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 l2 and PQ is not 90°, it follows that that angle must always be 90°.//
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.)