Centroid Theorem

Consider a triangle with vertices A, B, and C as well as two of its medians. Let G be the point of intersection of the medians.

Let Q be a point on AC such that QE is parallel to BD.

In the diagram, ∠ EQC and ∠ BDC are corresponding angles. Since EQ and BD are parallel, these two angles are congruent by the Corresponding Angles Theorem. The same is true for ∠ QEC and ∠ DBC.

Therefore, △ ECQ and △ BCD have two pairs of congruent angles and are similar by the Angle-Angle Similarity Theorem. Similar reasoning can be used to show that △ AGD and △ AEQ are also similar. △ ECQ~△ BCD △ AGD~△ AEQ By the definition of a median, E is the midpoint of BC, and therefore, E divides BC into two congruent segments. Note that congruent segments have equal lengths. This information and the Segment Addition Postulate imply that the length of BC is two times the length of EC. BE=EC BC=BE+EC ⇒ BC=2EC Therefore, the scale factor of the similar triangles is 12. That means DC=2QC. Furthermore, by the Segment Addition Postulate, DC=DQ+QC. Then, using the Transitive Property of Equality and the Subtraction Property of Equality the following is obtained.

Since QC and DQ are equal, Q is the midpoint of DC.

Remembering that AD=DC, the ratio of DQ to AD can be calculated. DQ/AD=DQ/DC=1/2 Note that corresponding parts of similar triangles are proportional. Therefore, since △ AGD and △ AEQ are similar, the ratio of DQ to AD is equal to the ratio of GE to AG. GE/AG=1/2 ⇔ GE=1/2AG This information can be used to express AG in terms of AE.

This means that AG is two-thirds of AE. Now, consider △ ABC and its medians CF and AE. Let K be the point of intersection of these medians.

Let R be a point on AB such that ER is parallel to CF.

By following the same reasoning as before, it can be proved that AK is two-thirds of AE. Therefore, G and K are the same points. That means the medians are concurrent — they meet at one point.

Before it was shown that AG= 23AE. By using similar arguments, it can be also shown that BG= 23BD and CG= 23CF.