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. Q.E.D. |
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