# Dissecting Triangles

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*medians,*

*midsegments,*and the

*centroid.*

## Perpendicular Bisector Theorem

Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.

This can be proven using congruent triangles.## Midsegment

A midsegment is a segment that connects the midpoints of two of the sides of a triangle.

Since a triangle has three sides, there are three midsegments.

## Triangle Midsegment Theorem

The segment that connects the midpoints of two sides of a triangle — a midsegment — is parallel to the third side of the triangle and half the length.

If $\overline{DE}$ is a midsegment of $\triangle ABC,$ $\overline{DE} \parallel \overline{AB}$ and $DE=\frac{1}{2}AB.$

## Triangle Proportionality Theorem

If a segment parallel to one of the sides in a triangle is drawn between the other sides, the segment divides the other two sides proportionally.

For $\triangle ABC,$ since $\overline{DE} \parallel \overline{AB},$ $\dfrac{AD}{DC}=\dfrac{BE}{EC}.$

This can be proven using congruent angles.## Median

A median of a triangle is a line segment between the midpoint of one side and the opposite vertex.

Since a triangle has three vertices, every triangle has three medians.

All medians intersect each other at one point, the centroid.## Centroid

The three medians of a triangle intersect at one point called the centroid.

## Centroid Theorem

The centroid of a triangle is two-thirds of the distance from each vertex to the midpoint of the opposite side.

If $\overline{AE},$ $\overline{BD},$ and $\overline{CF}$ are the medians of $\triangle ABC,$ then $AP=\tfrac{2}{3}AE, BP=\tfrac{2}{3}BD, \text{ and } CP=\tfrac{2}{3}CF.$

This can be proven using midpoints and parallel lines.

## Exercises

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