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?
11.2: Prove It: Parallel Segments
Does a line parallel to one side of a triangle always create similar triangles?
 Create several examples. Decide if the conjecture is true or false. If it’s false, make a more specific true conjecture.

Find any additional information you can be sure is true.
Label it on the diagram.  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.
 Segments \(AB\) and \(EF\) are parallel.
 Segments \(BD\) and \(FG\) are parallel. Segment \(EG\) is 12 units long. Segment \(EB\) is 2.5 units long.
Find the lengths of sides \(CE,CB\), and \(CA\) in terms of \(x, y,\) and \(z\). Explain or show your reasoning.
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 AngleAngle 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, \(\frac{BG}{GA}\)=\(\frac{BF}{FC}\).
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 \(A'B'C'\).