### MATH325: Lecture 8

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Last class we proved:

Lemma D: If line B'C' is parallel to line BC, B' is the midpoint of segment AB, and C' is on segment AC, then C' must be the midpoint of segment AC.

For homework, you proved the converse. These theorems are a very specific case of Theorem 1.7.1. (Theorem 1.7.1 is true for any similar pair of points B' and C', whereas Lemma D and its converse only hold for midpoints.)

Theorem 1.7.1: Let B' and C' be on the respective sides AB and AC of a triangle ABC. Then B'C' is parallel to BC if and only if:

|AB'|/|AB| = |AC'|/|AC|.

Discussion of proof of 1.7.1: The "if and only if" in the theorem means that we have to prove that each half of the theorem implies the other. We must prove both a statement and its converse in order to complete the proof of this one theorem. If we can prove that when line B'C' is parallel to line BC the side lengths of ABC and AB'C' are proportional it is easy to prove the converse; unfortunately I do not know a good way to prove the first statement.

We can prove the statement in the special case in which which the ratio |AB'|/|AB| is 1/2 -- Lemma D tells us that if B' and C' are on the respective sides AB and AC of a triangle ABC and B'C' is parallel to BC, then |AB'|/|AB| = |AC'|/|AC| = 1/2. If we could prove this statement for ratios other than 1/2 we would have proven half of our theorem:

Theorem 1.7.2: Let B' and C' be on the respective sides AB and AC of a triangle ABC. If B'C' is parallel to BC then:

|AB'|/|AB| = |AC'|/|AC|.

In Chapter 9 the author proves this by first proving it's true for ratios equal to any fraction p/(2n) and then applying a limiting argument something like the comparison test from Calculus III. The idea is that because you can find a number p/(2n) as close to |AB'|/|AB| as you like, you can "trap" the line B'C' between two lines that are parallel to BC. You can move those lines as close together as you like with B'C' still trapped between them, so B'C' must be parallel to BC.

The proof that 1.7.1 is true for ratios equal to p/(2n) is probably just a repeated version of our proof of Lemma D.

Although we know that around 200 BC Archimedes used a limiting process (the "method of exhaustion") to estimate the area of a circle, this is not the technique that Euclid used to prove his version of Theorem 1.7.1.

In any case, once we've proven that if B'C' is parallel to BC then the sides of ABC are proportional to the sides of A'B'C', we can quickly use that to prove the converse:

Theorem 1.7.3: Let B' and C' be on the sides AB and AC of a triangle ABC. Suppose

|AB'|/|AB| = |AC'|/|AC|.

Then B'C' is parallel to BC.

Proof: Let C'' be the point of intersection of segment AC and the line through B' that is parallel to BC. Then Theorem 1.7.2 tells us that because B'C'' is parallel to BC, it must be true that:

|AB'|/|AB| = |AC''|/|AC|.

But we started with the assumption that:

|AB'|/|AB| = |AC'|/|AC|.

And so a simple algebraic argument concludes that |AC'| = |AC''|; thus C' = C'' and B'C' is the same line as B'C'', and so is parallel to BC.// Definition 10: Two triangles ABC and A'B'C' are similar if their corresponding angles are congruent. Corollary 1.7.4: If triangle ABC is similar to triangle A'B'C', then

|A'B'|/|AB| = |A'C'|/|AC| = |B'C'|/|BC|.

Discussion of proof: To prove this, use an isometry to "move" the smaller triangle into the larger one. Then apply Theorem 1.7.1. I believe Euclid's proof of Theorem 1.7.1 is based on a comparison of the triangles' areas. Although the author has already discussed the Pythagorean Theorem in section 1.1, he hasn't yet defined area. He's assuming that: