Given: In the figure alongside, B and F are the midpoints of sides AC and AE. AG = GD.
R.T.P.: H is the center of CE, therefore AH is the third median of triangle ACE, therefore the medians of a triangle are concurrent.
Proof: In triangle ACD, B and G are the midpoints of sides AC and AD.
Therefore, BG//CD and BG = half the length of CD.
In the same way it can be proven that FG//DE and FG = half of the length of DE.
Since CG//DE and CD//GE, CDEG is a parallelogram.
This makes CH=HE since the diagonals of a parallelogram bisect each other.
Therefore AH is the third median of the triangle ACE.
Therefore G, the point of intersection of the three medians is the point of concurrency.
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