Problem
GEO-B1-M06-P027 Area Form of Ceva's Theorem
#27
★★★★★ Level 5 of 5
In triangle \(ABC\), lines \(AD\), \(BE\), \(CF\) meet at one point \(P\), where \(D\in BC\), \(E\in CA\), \(F\in AB\). Prove that \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\).
Express each segment ratio through area ratios of triangles with vertex \(P\).
Since \(P,A,D\) are collinear, triangles \(ABP\) and \(PBD\) have heights to this line from \(B\), and triangles \(ACP\) and \(PCD\) have heights from \(C\). Comparing fractions on the same line gives \(\frac{BD}{DC}=\frac{S_{ABP}}{S_{ACP}}\). Similarly, \(\frac{CE}{EA}=\frac{S_{BCP}}{S_{ABP}}\) and \(\frac{AF}{FB}=\frac{S_{ACP}}{S_{BCP}}\). Multiplying, we get \(1\).
This is a strong preparatory problem for Book 2; here it is presented as an advanced area chasing example.