Problem
GEO-B1-M08-P028 The Third Median Through Areas
#28
★★★★★ 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.
Since \(G\) lies on the median from \(A\), \(S_{GAB}=S_{GAC}\). Since \(G\) lies on the median from \(B\), \(S_{GAB}=S_{GBC}\). Hence \(S_{GAC}=S_{GBC}\). These triangles have common base \(GC\), so points \(A\) and \(B\) are equally distant from line \(GC\). Therefore \(GC\) passes through the midpoint of \(AB\), that is, \(AF=FB\).
A strong problem proving a median property through areas.