Problem
GEO-B2-M10-P019 Ceva Through Areas
#19
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D,E,F\) lie on \(BC,CA,AB\). Given \([ABD]:[ACD]=2:3\), \([BCE]:[BAE]=3:4\), \([CAF]:[CBF]=2:1\). Prove that \(AD,BE,CF\) are concurrent.
Convert area ratios into segment ratios on the sides.
Since each corresponding pair of triangles has a common height, we get \(BD:DC=2:3\), \(CE:EA=3:4\), \(AF:FB=2:1\). Then \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{4}\cdot 2=1\). By Ceva, lines \(AD,BE,CF\) are concurrent.
Here the method has two steps: areas, then Ceva.