Problem
GEO-B1-M03-P016 Medians and the Ratio \(2:1\)
#16
★★★☆☆ Level 3 of 5
In triangle \(ABC\), medians \(BM\) and \(CN\) meet at \(G\). Prove that \(BG:GM=CG:GN=2:1\).
Take the midpoints of \(BG\) and \(CG\), then use midlines.
Let \(P\) and \(Q\) be the midpoints of \(BG\) and \(CG\). Then \(PQ\parallel BC\) and \(PQ=\frac{1}{2}BC\). Since \(M\) and \(N\) are the midpoints of \(AC\) and \(AB\), segment \(MN\) is also parallel to \(BC\) and equals \(\frac{1}{2}BC\). Hence \(PQ\parallel MN\) and \(PQ=MN\), so \(MNPQ\) is a parallelogram. Its diagonals \(MP\) and \(NQ\) meet at \(G\) and bisect each other. Therefore \(GM=GP\), and since \(P\) is the midpoint of \(BG\), \(BP=PG\). Thus \(BG=2GM\). Similarly, \(CG=2GN\).
This proof is a bridge to more systematic work with medians.