Problem
GEO-B2-M06-P024 No Circles Needed
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\) are chosen arbitrarily. Line \(DE\) meets the extension of \(AB\) at \(X\). Point \(F\) is chosen on side \(AB\) so that \(AF:FB=AX:XB\). Prove that if \(AD\) and \(BE\) meet at \(P\), then points \(C,P,F\) are collinear.
Menelaus gives the product for \(D,E,X\), then Ceva gives concurrence of \(AD,BE,CF\).
By Menelaus for collinear \(D,E,X\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AX}{XB}=1\). By the condition, \(\frac{AX}{XB}=\frac{AF}{FB}\), so \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). By Ceva, lines \(AD\), \(BE\), \(CF\) are concurrent. But \(AD\) and \(BE\) already meet at \(P\), so the third line \(CF\) also passes through \(P\). Therefore \(C,P,F\) are collinear.
A strong final problem: the student must assemble Menelaus and Ceva in the correct order.