# Proportionality Relationships

## Big Picture

Proportionality are important in many relationships. The fundamental properties of proportions are useful in geometric proofs.

## Key Terms

Midsegment: The segment that joins the midpoints of two sides of a triangle.

Transversal: A line that intersects a set of lines (may or may not be parallel).

Angle Bisector: A ray that divides an angle into two congruent angles, each with a measure equal to exactly half of the original angle.

## Triangle Proportionality

Recall that every triangle has three midsegments.

• Midsegment Theorem: The midsegment of a triangle is parallel to one side of a triangle and divides the other two sides in half.

The midsegment divides the other two sides of the triangle proportionally.

• The ratio for the triangle below is a : a or b : b, which both simplify to a ratio of 1:1.
• Creates two similar triangles

The Midsegment Theorem is a special case of the Triangle Proportionality Theorem.

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides, then it divides those sides into proportional segments.

• If $$\overline{XY} || \overline{DF}$$, then $$\frac{EX}{XD} = \frac{EY}{YF}$$

The converse is also true.

Converse of the Triangle Proportionality Theorem: If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

## Parallel Lines and Transversals

Theorem: If two transversals intersect the same set of parallel lines, then the parallel lines divide the transversals into proportional segments.

• $$\frac{AB}{BC}$$ = $$\frac{DE}{EF}$$

This theorem can be expanded to any number of parallel lines and any number of transverals.

• All the corresponding segments of the transversals are proportional.

## Angle Bisector

Theorem: If a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the lengths of the other two sides.

• $$\overrightarrow{AC}$$ bisects $$\angle BAD$$ so that $$\angle BAC \cong CAD$$, so $$\frac{BC}{CD}$$ =  $$\frac{AB}{AD}$$.