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If AE, BD, and CF are the medians of △ABC, then the following statements hold true.
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.GE=21AG
a=22⋅a
b1⋅a=ba
Add fractions
LHS⋅2=RHS⋅2
LHS/3=RHS/3
ca⋅b=ca⋅b
Rearrange equation
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=32AE. By using similar arguments, it can be also shown that BG=32BD and CG=32CF.