Problem
GEO-B2-M06-P018 One Pair of Points, Two Theorems
#18
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\). Lines \(AD\) and \(BE\) meet at \(P\), \(CP\) meets \(AB\) at \(F\), and \(DE\) meets the extension of \(AB\) at \(X\). Prove that \(\frac{AF}{FB}=\frac{AX}{XB}\).
Write Ceva for \(D,E,F\) and Menelaus for \(D,E,X\).
By Ceva, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). By Menelaus for line \(DEX\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AX}{XB}=1\). The first two factors are the same, so the third factors are equal: \(\frac{AF}{FB}=\frac{AX}{XB}\).
A strong problem on applying both theorems at once.