Given any triangle ABC, let AE, BG, and CF be the medians (figure 4). Ceva's Theorem states that (AF/FB)(BE/EC)(CG/GA)=1 if and only if the cevians are concurrent. Since CF is a median, then AF=FB. Similarly, BE=EC, and CG=GA. So, AF/FB=1, BE/EC=1, and CG/GA=1. Substituting for the values, we get (1)(1)(1)=1. Therefore, the medians are concurrent.

Figure 4

 

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