Problem
GEO-B1-M06-P029 The Third Median Through Areas
#29
★★★★★ Level 5 of 5
In triangle \(ABC\), the medians from \(A\) and \(B\) meet at point \(G\). Line \(CG\) meets \(AB\) at point \(F\). Prove that \(AF=FB\).
A point on a median gives two equal areas. Apply this to two medians.
Since \(G\) lies on the median from \(A\), we have \(S_{GAB}=S_{GAC}\). Since \(G\) lies on the median from \(B\), we have \(S_{GAB}=S_{GBC}\). Hence \(S_{GAC}=S_{GBC}\). Triangles \(GAC\) and \(GBC\) have common base \(GC\), so points \(A\) and \(B\) are at equal distances from line \(GC\). Therefore line \(GC\) passes through the midpoint of \(AB\), that is, \(AF=FB\).
A neat area proof of concurrence of medians.