Lesson 11

Splitting Triangle Sides with Dilation, Part 2

  • Let’s investigate parallel segments in triangles.

11.1: Notice and Wonder: Parallel Segments

What do you notice? What do you wonder?

\(\overleftrightarrow {AC} \parallel \overleftrightarrow{MN}\)

Parallel lines A C and M N slope slightly down from left to right. Point B is below the lines. Segment B A intersects M N at M and A C at A. Segment B C intersects M N at N and AC at C. 

11.2: Prove It: Parallel Segments

Does a line parallel to one side of a triangle always create similar triangles?

  1. Create several examples. Decide if the conjecture is true or false. If it’s false, make a more specific true conjecture.
  2. Find any additional information you can be sure is true.
    Label it on the diagram.

    \(\overleftrightarrow {AC} \parallel \overleftrightarrow{MN}\)

    Parallel lines A C and M N slope slightly down from left to right. Point B is below the lines. Segment B A intersects M N at M and A C at A. Segment B C intersects M N at N and AC at C. 
  3. Write an argument that would convince a skeptic that your conjecture is true.

11.3: Preponderance of Proportional Relationships

Find the length of each unlabelled side.

  1. Segments \(AB\) and \(EF\) are parallel.

    \(\overline{AB} \parallel \overline{EF}, \overline{AD} \perp \overline{DB}\)

    Triangle D E F with horizontal side D F.
  2. Segments \(BD\) and \(FG\) are parallel. Segment \(EG\) is 12 units long. Segment \(EB\) is 2.5 units long.

    \(\overline{BD} \parallel \overline{FG}\)

    Triangle E F G with horizontal side F G.


Find the lengths of sides \(CE,CB\), and \(CA\) in terms of \(x, y,\) and \(z\). Explain or show your reasoning.

Diagram of right triangle.

 

Summary

In triangle \(ABC\), segment \(FG\) is parallel to segment \(AC\). We can show that corresponding angles in triangle \(ACB\) and triangle \(GFB\) are congruent, so the triangles are similar by the Angle-Angle Triangle Similarity Theorem. There must be a dilation that sends triangle \(GFB\) to triangle \(ACB\), and so pairs of corresponding side lengths are in the same proportion. Then we can show that segment \(GF\) divides segments \(AB\) and \(CB\) proportionally. In other words, \(\frac{BG}{GA}\)=\(\frac{BF}{FC}\).

\(\overleftrightarrow{FG} \parallel \overleftrightarrow{AC}\)

Triangle A B C. Point G on segment A B. Point F on segment B C. Directed line segments A C and G F drawn with arrows point towards Points C and F.
 

For example, suppose \(G\) is \(\frac23\) of the way from \(A\) to \(B\) and \(F\) is \(\frac23\) of the way from \(C\) to \(B\). Then if \(BA=9\) and \(BC=12\), we know that \(GA=6\) and \(FC=8\). What will \(BG\) and \(BF\) equal? Since \(BG=3\) and \(BF=4\), we know that \(\frac36=\frac48\) and can show that \(\frac{BG}{GA}\)=\(\frac{BF}{FC}\).

This argument holds in general. A segment in a triangle that is parallel to one side of the triangle divides the other 2 sides of the triangle proportionally.

Glossary Entries

  • similar

    One figure is similar to another if there is a sequence of rigid motions and dilations that takes the first figure onto the second.

    Triangle \(A'B'C'\) is similar to triangle \(ABC\) because a rotation with center \(B\) followed by a dilation with center \(P\) takes \(ABC\) to \(A'B'C'\).

    Triangle ABC, rotated and then dilated.