MATH325: Lecture 7

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Similar Triangles

There are too many theorems in geometry and too many ways of understanding those theorems to cover them all in one class. The Star Trek and Carpenter's lemmas lead the way to many interesting properties of angles and arcs, but we won't take the time to pursue those topics right now.

Section 1.7 introduces the concept of similar triangles, another tool commonly seen in geometry classes. We won't prove theorem 1.7.1 completely, but we'll try to get an idea of the process the author uses to prove it.

Extra credit: Describe how Euclid proves Theorem 1.7.1.

Discussion of Similar Triangles: The study of geometry arose not through some intellectual fascination with axiomatic systems, but because people wanted to measure things. Applications of geometry range from surveying through navigation to astronomy. One important tool used in these applications is the notion of similar triangles -- the entire subject of trigonometry is based on facts about similar triangles.

A common high school activity involving similar triangles is estimating the height of a building or tree based on measurements taken at ground level. I've looked up web sites that include a brief description of this activity and a lesson plan.

Theorem 1.7.1 lays the groundwork for the definition of similar triangles and introduces the topic of proportionality. We won't prove it completely, but we will try to understand why it's true and what it implies.

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|.

Notice that this is the first place where we really compare different lengths or distances. So far, the lengths we've worked with -- have been equal, doubled, halved, or added. This theorem introduces the idea of looking at the ratios of two lengths. If we only talk about how lengths are related to other lengths, we never need to worry about whether we're using metric or US customary units (or degrees or radians). If imprecise measures like hands, paces, and fistfuls were the standard units in Euclid's time, that might explain why ratios were so popular in precise mathematical calculations. This also explains why "ruler and compass" constructions are traditionally done with "straight edge and compass".

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 A'B'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: 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. Proof of Lemma D: We aim to prove that AC' is congruent to C'C by proving that triangle 1 is congruent to triangle 4. (See diagram above.)

By Axiom 5 we can construct a line parallel to AC through point B'. Let A' be the point of intersection of that line and edge BC.

Step 1: Triangle 1 is congruent to triangle 2.

BB' is congruent to B'A because B' is a midpoint.
Angle AB'C' is congruent to angle B'BA' because B'C' is parallel to BC.
Angle AC'B' is congruent to angle B'A'B because B'A' is parallel to AC.
Therefore triangle AC'B' is congruent to triangle B'A'B by angle, angle, side.

Step 2: Triangle 3 is congruent to triangle 1.

C'B' is congruent to B'C'.
Angle AC'B' is congruent to angle A'B'C' because B'A' is parallel to AC.
A'B' is congruent to AC' because triangle 2 is congruent to triangle 1.
Therefore triangle AC'B' is congruent to triangle A'B'C' by side, angle, side.

Step 3: Triangle 4 is congruent to triangle 1.

Angle B'C'A is congruent to angle A'CC' because B'C' is parallel to BC.
A'C' is congruent to AB' because triangles 1 and 3 are congruent
Angle C'A'C is congruent to angle AB'C' because supplementary angles sum to 180°.
(By congruent triangles, m∠B'AC' = m∠C'A'B'. Denote this angle size by α. Similarly, let β = m∠BA'B' = m∠B'C'A. Then because the angles are supplementary, m∠C'A'C + α + β = 180° and m∠AB'C' + β + α = 180°.)
Therefore, by angle, angle, side triangle B'C'A is congruent to triangle A'CC'.

Because triangle B'C'A is congruent to triangle A'CC', C'A is congruent to CC', and so C' must be the midpoint of segment AC.//

Discussion of proof of converse of Lemma D: For homework, you were asked to prove that if B' and C' are midpoints of the edges (if the ratios above are 1/2) then B'C' is parallel to BC. If you're stuck on the proof, try starting with a line through B' parallel to BC and proving that C' must lie on that line. Alternately, assume that C' does not line on the line through B' that is parallel to BC and derive a contradiction.