We also proved some familiar theorems about parallel lines and
alternate interior angles. For example, Corollary 1.4.4 tells us that
if the alternate interior angles are equal where a transversal l
crosses lines l_{1} and l_{2}, then l_{1} and
l_{2} must be parallel.

Is the converse of this true? In other words, if l_{1} and
l_{2} are parallel, can we show that a transversal line l will
always have equal alternate interior angles where it meets
l_{1} and l_{2}? The answer is "yes".

**Lemma B:** If l_{1} and l_{2} are any pair of
parallel lines, and if line l intersects l_{1} and
l_{2}, then the alternate interior angles formed by those
intersections are equal.

**Proof:** Let's call the angle between l_{1} and
l_{2} where l meets l_{1} angle α. Call β the
angle on the same side of l where it meets l_{2}.

Because l_{1} and l_{2} are parallel, Euclid's fifth
postulate tells us that the measure of α plus the measure of
β must equal 180°.

Because the sum of the measures of supplementary angles is 180°, we know that the two angles adjacent to β have the same measure as angle α. This completes the proof of the lemma -- the alternate interior angles are equal.//

(The text does not prove this as a separate lemma, but it is a useful result in its own right. Now we can refer to it as "Lemma B" rather than having to say "we can use the fifth postulate to prove" or "by a similar argument to the one used in Theorem 1.4.6". Also, separating this proof out from the proof of Theorem 1.4.6 may make the proof of 1.4.6 easier for our students to understand.)

Lemma B should be familiar from previous geometry classes. The following theorem should also be familiar:

Did you ever wonder how we know this is true? Combining Euclid's fifth postulate with Lemma B makes this easy to prove!

Here's a nice way to introduce the proof: have everyone in your class cut out a triangle (or hand out an assortment of triangles). Then rip the points off the triangle and place them on a desk next to each other. The three "pie slices" ripped off the corners of the triangle fill out half a pie!

This sort of experiment is very convincing to students, but shouldn't be mistaken for a proof. One of the nice things about mathematics is that we can say "I know for sure that it's true" rather than thinking "it was true every time I checked it, but maybe next time I'll find a counterexample."

**Definition 9: Exterior Angle.** The *exterior angle at* A in
triangle ABC is one of the angles adjacent to angle BAC at the
intersection of line AB and line AC. (Since the two angles at that
intersection are equal, it doesn't matter which we choose.)

**Proof of Theorem 1.4.6:** Suppose we have triangle ABC. We want
to show that the sum of the measures of the angles in this triangle is
180°.

Using Axiom 5 we can construct a line l through A parallel to line BC. There are three angles formed where segments BA and AC meet line l; call them β, ∠BAC, and γ as shown in figure 1.10.

By Lemma B, the measure of angle ABC equals the measure of angle β. Similarly, angle γ is congruent to angle ACB.

Since β, ∠BAC and γ add up to 180°, the sum of the angles of the triangle must also be 180°.//

Homework: Exercise 1.23 (explain your reasoning, but a formal proof is not required)

Our textbook uses Side-Side-Side to prove this theorem very quickly. As always, make yourself a sketch so that you can follow the proof.

**Theorem 1.5.1 (Pons Asinorum):** The base angles of an isoceles triangle are equal.

**Proof:** We know we're starting with an isosceles triangle ABC;
let's assume |AB| = |AC|. Notice that |BC| = |BC|, so corresponding
sides of ΔABC and ΔACB are congruent.

Then by Side-Side-Side (Theorem 1.3.1), triangle ABC is congruent to triangle ACB.

Therefore angle ABC must be congruent to angle ACB.//

The textbook goes a little further than is needed. The author's remark that the isometry taking ABC to ACB must be a reflection makes the proof more concrete and is a nice reminder of the definition of congruence.

Exercises 1.24 and 1.25 are marked with daggers (†). This means that they will be used later in the textbook. Exercise 1.24 is pretty easy to prove (hint: use Exercise 1.14).

Homework: Formally prove Exercise 1.24. You may assume that the result of Exercise 1.14 (ASA) is true.

**Theorem 1.6.1 (Star Trek Lemma):** The measure of an inscribed
angle is half the angular measure of the arc it subtends.

There are three lessons to be learned here. First, that the measure of an angle inscribed in a circle can be determined by measuring an angle at the center of the circle. Second, that mathematicians are allowed to make up names for their theorems to help remember them. And third, a lesson from my own experience: theorems like this are in the text book for a reason. When I learned this theorem (before it got its new name) I did not see its usefulness and made little effort to remember it. However, I have needed it many times since then and often regretted not paying closer attention when this theorem was first presented to me.

**Discussion of proof:** Mathematicians are lazy -- they don't like
to repeat work that has already been done. This is another reason why
lemmas are useful -- we can re-use them to prove other theorems later.

To completely prove the Star Trek Lemma one must repeat the proof four times for four different cases. The author is lazy and leaves three of those proofs to us to do for homework. I'm lazy, so two of them aren't even assigned for homework, although it's possible one case could appear on Test 1.

**Proof** of the theorem for the case that gave it its name:

In the case we will consider, angle BAC is acute and the center O of the circle lies inside this angle. The diagram looks like the one in Figure 1.11, which in turn looks a little bit like the Starfleet Insignia from Star Trek.

Construct radii OA, OB and OC. These segments all have equal length.

Extend segment OA until it meets the circle at point D. (Axiom 2 says we can extend the segment. Chapter 9 says that it must meet the circle.) Segment OD is also a radius and so congruent to the other radii.

We now have a whole lot of isosceles triangles, we know that the sum of the measures of supplementary angles is 180°, and we know that the sum of the angles of a triangle sum to 180°. By examining the correct triangles and angle sums, we can eventually compare angle BAC to angle BOC.

Begin with isosceles triangle AOB. We know that angles ABO and BAO are congruent, and that the sum of the angles of that triangle is 180°. Then:

m ∠ABO + m ∠BAO + m ∠ BOA = 180°

Replacing m ∠ABO with m ∠BAO (because the two are equal), we get:

2 * m ∠BAO + m ∠BOA = 180°

But because the angles are supplementary, we also know that:

m ∠DOB + m ∠BOA = 180°

Comparing the two equations we conclude that: m ∠DOB = 2 * m ∠BAO.

Try to reproduce this argument to prove to yourself that:

m ∠DOC = 2 * m ∠CAO.

We've proven that the inscribed angle is half the angle at the center of the circle for both "sides" of the inscribed angle. Add these results together to get the conclusion we're trying to reach:

m ∠BOD + m ∠DOC = 2 * (m ∠BAO + m ∠OAC)

m ∠BOC = 2 * m ∠BAC.//

We just proved this theorem for the case where the center of the circle lies inside the inscribed angle. There are two cases we won't prove it for: the center is to one side of the angle, and the center is opposite the (obtuse) angle. In both cases it is still true, and the proof is similar.

The third case is the one in which triangle BOC is degenerate -- in other words, BC is a straight line and BAC is a right angle. The fact that when a right angle is inscribed in a circle, the hypotenuse is a diameter of the circle is the key to the "Carpenter's Construction" of a circle -- an easy to describe practical application of geometry. We will prove this special case of the Star Trek lemma in the next lecture.