Problem
GEO-B2-M06-P015 Proof of Ceva by Areas
#15
★★★★☆ Level 4 of 5
Let in triangle \(ABC\), lines \(AD\), \(BE\), \(CF\) be concurrent at \(P\). Prove that \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\).
Express ratios on sides through areas of triangles with a common altitude.
Since triangles \(PBD\) and \(PDC\) have a common altitude to \(BC\), \(\frac{BD}{DC}=\frac{[PBD]}{[PDC]}\). Similarly, \(\frac{CE}{EA}=\frac{[PCE]}{[PEA]}\), \(\frac{AF}{FB}=\frac{[PAF]}{[PFB]}\). Grouping the areas around point \(P\), the product telescopes and equals \(1\). This is the necessary part of Ceva's theorem.
This is not just an application, but understanding where the formula comes from.