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Medians, the centroid and the midsegment

The three medians meet at the centroid G, which cuts each median 2 : 1 from the vertex; the segment joining two midpoints is parallel to the third side and half as long.

High school Lesson 7 of 9 Interactive figure Practice

First, the rules

Where the medians meet

Median

A segment from a vertex to the midpoint of the opposite side. Every triangle has three; they always meet at one point, the centroid G.

★ The 2 : 1 rule

Examples 1, 2

|AG| = 2 |GD|, so |AG| = 23|AD| and |GD| = 13|AD|
G cuts each median two thirds of the way from the vertex.

A median halves the area

The two triangles on either side of a median have equal area (same height, equal bases). The three medians cut the triangle into six equal areas.

★ Midsegment theorem

Example 1

The segment joining the midpoints of two sides is parallel to the third side and half its length.

Midsegment meets the median at its midpoint

Example 1

If F and E are the midpoints of AB and AC, then FE crosses the median AD exactly at the midpoint of AD.

In a right triangle

The median to the hypotenuse is half the hypotenuse: the midpoint of the hypotenuse is equidistant from all three vertices.

See it

Drag a corner — G keeps the 2 : 1

All three medians and the centroid are drawn. Whatever shape you make, each median is cut 2 : 1. Switch on the midsegment to see it stay parallel to the base and exactly half as long.

Step by step

Worked examples

True or false

centroid · midsegment

In triangle ABC, D is the midpoint of BC, E the midpoint of AC and F the midpoint of AB. AD meets FE at K and the medians meet at G. Decide:
I. |AD| = 3 |GD| · II. |AD| = 6 |KD| · III. |BD| = 2 |KE|

Ask yourself: F and E are midpoints. Which side is FE parallel to, and how long is it?

Show the solution

Background: FE is a midsegment, so FE ∥ BC and FE cuts the median AD at its midpoint: |AK| = |KD|.
I. G divides the median 2 : 1: |AG| = 2|GD|, so |AD| = 3|GD|. True.
II. K is the midpoint of AD, so |AD| = 2|KD|, not 6|KD|. False.
III. In triangle ADC, K and E are midpoints, so KE is a midsegment: |KE| = |DC|2 = |BD|2, i.e. |BD| = 2|KE|. True.

Answer I. true · II. false · III. true

Perimeter of triangle AGB

the 2 : 1 rule

In triangle ABC the medians BE and AD meet at G. |BE| = |AB| = 24 cm and |AD| = 18 cm. Find the perimeter of triangle AGB.

Schematic, not to scale.

Ask yourself: In what ratio does G cut each median? Which part of 24 and of 18 is on the vertex side?

Show the solution

Median BE, cut in three parts of 24 / 3 = 8: |BG| = 2 · 8 = 16, |GE| = 8.
Median AD, three parts of 18 / 3 = 6: |AG| = 2 · 6 = 12, |GD| = 6.
Sides of AGB: |AB| = 24, |BG| = 16, |AG| = 12.
24 + 16 + 12

Answer 52 cm

Practice: medians and midsegments

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A median length, a piece of it, or a side — find what is asked using the 2 : 1 rule and the midsegment theorem. Your best score is kept in this browser only.

About this lesson

Medians and the centroid: the three medians of a triangle meet at one point G that divides each median in the ratio 2 : 1 measured from the vertex. The midsegment theorem: the line joining the midpoints of two sides is parallel to the third side and half its length. Drag the triangle, read the lengths, then practise.

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