MATH325: Lecture 9


Syllabus | Schedule | Grading Keys/Rubrics

Similar Triangles

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|.

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. For this class, let's use the following definition:

Definition: The area of a (planar) triangle is 1/2 * base * height.

Of course, this raises the point that we haven't discussed or defined lengths yet, either.

Section 1.9

Section 1.9 contains a nice application of Theorem 1.7.1. Also, it is common when studying triangles to look at lines like medians, altitudes and angle bisectors and prove theorems about how those lines meet. This is one example of such a theorem.

Definition: In triangle ABC, the line segment joining A to the midpoint of segment BC is a median of the triangle.

Theorem 1.9.1: The three medians of a triangle intersect at a common point G, which we call the centroid of ABC. In addition, if A', B' and C' are the midpoints of BC, AC and AB, respectively, then:

|A'G|/|AG| = |B'G|/|BG| = |C'G|/|CG| = 1/2.

Discussion of Proof: We let G be the intersection of segments BB' and CC' and then let G' be the intersection of segments AA' and BB'. We'll use Theorem 1.7.1 to prove:

|B'G|/|BG| = |B'G'|/|BG'| = 1/2

(if we didn't have to prove this in order to prove the theorem, it probably wouldn't be part of the theorem statement). From this we will conclude that G = G' and so all three medians meet at a single point.

Proof: Not surprisingly, we use similar triangles and corollary 1.7.4 to complete the proof.

Start by drawing segments B'C', BB', and CC'. Let G be the intersection of segments BB' and CC'. Angles C'GB' and BGC are opposite angles, so are congruent. Angles B'C'G and BCG are alternate interior angles as are angles GB'C' and GBC. Consequently, corresponding angles of triangles GB'C' and GBC are congruent and so those two triangles must be similar.

By applying Theorem 1.7.1 and corollary 1.7.4 we can show that |C'B'|/|BC| = 1/2. (If you're not sure why this is true, go through the steps of proving AC'B' and ABC are similar and therefore the ratios of their edge lengths are 1 to 2.) Since triangles BB'C' and B'BC are similar and |C'B'|/|BC| = 1/2, it must be true that |C'G|/|CG| = |B'G|/|BG| = 1/2.

Next we construct segments AA' and BB' and name their point of intersection G'. An identical proof shows that |A'G'|/|AG'| = |B'G'|/|BG'| = 1/2.

In particular, |B'G|/|BG| = |B'G'|/|BG'| and so |B'G| = |B'G'|. From this we can conclude that G = G'. (There are only two points distance |B'G| away from B' on line BB', and one of them is outside the triangle. Since G and G' on line BB' and both inside the triangle, they must be the same point.)//

The altitudes of a triangle are lines through the vertices perpendicular to the opposite edges. Is it true that the altitudes of a triangle must always meet at the same point? If so, what can you find out about that point?

Do the angle bisectors of the three internal angles of the triangle always meet at the same point? If so, what can you find out about that point?