#6 Ceva and Menelaus

By Ceva, \(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{AF}{FB}=1\). The first factors give \(\frac{1}{2}\), so \(\frac{AF}{FB}=2\). Answer: \(AF:FB=2:1\).

Compute: \(\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{3}{2}=1\). Therefore, by Ceva's theorem, lines \(AD\), \(BE\), \(CF\) are concurrent.

Let \(D,E,F\) be the midpoints of \(BC,CA,AB\). Then \(\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=1\). The product is \(1\), hence by Ceva the medians \(AD\), \(BE\), \(CF\) are concurrent.

By Menelaus, \(\frac{2}{5}\cdot\frac{5}{3}\cdot\frac{AF}{FB}=1\). We get \(\frac{2}{3}\cdot\frac{AF}{FB}=1\), hence \(AF:FB=3:2\).

We have \(\frac{2}{3}\cdot\frac{3}{4}\cdot 2=1\). Exactly one point lies on an extension of a side, so by the positive form of Menelaus, points \(D,E,F\) are collinear.

By Ceva's theorem, the product being \(1\) for points on the sides of a triangle is equivalent to concurrence of the cevians. Therefore \(AD\), \(BE\), \(CF\) meet at one point.

By Ceva, \(\frac{3}{2}\cdot\frac{5}{6}\cdot\frac{AF}{FB}=1\). The first factors equal \(\frac{5}{4}\), hence \(\frac{AF}{FB}=\frac{4}{5}\). Answer: \(4:5\).

By Ceva, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AM}{MB}=1\). Thus \(2\cdot\frac{CE}{EA}\cdot 1=1\), so \(\frac{CE}{EA}=\frac{1}{2}\). Answer: \(CE:EA=1:2\).

With one external point, use the positive form: \(\frac{AD}{DB}\cdot\frac{BF}{FC}\cdot\frac{CE}{EA}=1\). We have \(\frac{2}{3}\cdot\frac{BF}{FC}\cdot\frac{1}{4}=1\). Hence \(\frac{BF}{FC}=6\), so \(BF:FC=6:1\).

By Menelaus, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Thus \(\frac{3}{4}\cdot\frac{CE}{EA}\cdot\frac{7}{2}=1\). We get \(\frac{21}{8}\cdot\frac{CE}{EA}=1\), hence \(CE:EA=8:21\).

Let the angle bisectors from \(A,B,C\) meet the opposite sides at \(D,E,F\). Then \(\frac{BD}{DC}=\frac{AB}{AC}\), \(\frac{CE}{EA}=\frac{BC}{BA}\), \(\frac{AF}{FB}=\frac{CA}{CB}\). The product of these three ratios is \(1\). By Ceva, the angle bisectors are concurrent.

By Ceva, \(\frac{AD}{DB}\cdot\frac{BF}{FC}\cdot\frac{CE}{EA}=1\). Here \(\frac{AD}{DB}=\frac{1}{2}\), and \(\frac{CE}{EA}=\frac{3}{2}\). Thus \(\frac{1}{2}\cdot\frac{BF}{FC}\cdot\frac{3}{2}=1\). Hence \(\frac{BF}{FC}=\frac{4}{3}\). Answer: \(BF:FC=4:3\).

We have \(\frac{AD}{DB}\cdot\frac{BF}{FC}\cdot\frac{CE}{EA}=1\). Thus \(\frac{3}{2}\cdot\frac{BF}{FC}\cdot\frac{1}{5}=1\), so \(\frac{BF}{FC}=\frac{10}{3}\). Answer: \(BF:FC=10:3\).

The product is \(\frac{AB}{AC}\cdot\frac{BC}{BA}\cdot\frac{CA}{CB}=1\). By Ceva's theorem, lines \(AD\), \(BE\), \(CF\) are concurrent.

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.

Let the directed distances from \(A,B,C\) to line \(l\) be \(h_A,h_B,h_C\). Since \(D\) lies on both \(BC\) and \(l\), point \(D\) divides \(BC\) in the ratio of directed distances to \(l\): \(\frac{BD}{DC}=-\frac{h_B}{h_C}\). Similarly, \(\frac{CE}{EA}=-\frac{h_C}{h_A}\) and \(\frac{AF}{FB}=-\frac{h_A}{h_B}\). Multiplying gives \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=-1\).

By Ceva, \(\frac{2}{3}\cdot\left(-\frac{3}{5}\right)\cdot\frac{AF}{FB}=1\). The first two factors give \(-\frac{2}{5}\). Hence \(\frac{AF}{FB}=-\frac{5}{2}\). The negative sign means that \(F\) lies on an extension of \(AB\).

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: \(\frac{AF}{FB}=\frac{AX}{XB}\).

By Ceva, \(\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{AF}{FB}=1\), so \(\frac{2}{5}\cdot\frac{AF}{FB}=1\), and \(AF:FB=5:2\). By Menelaus for \(D,E,X\): \(\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{AX}{XB}=1\), hence \(AX:XB=5:2\).

By Menelaus for line \(D E X\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AX}{XB}=1\). By the condition, \(\frac{AX}{XB}=\frac{AF}{FB}\), so \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). By Ceva's theorem, lines \(AD\), \(BE\), \(CF\) are concurrent.

Since \(D\) and \(E\) are internal, lines \(AD\) and \(BE\) meet inside the triangle, so line \(CP\) meets side \(AB\) at an internal point \(F\). A line through two internal points \(D\) and \(E\) meets the third side \(AB\) on its extension, so \(X\) is external. By Ceva, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\), and by Menelaus for \(D,E,X\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AX}{XB}=1\). Therefore \(\frac{AF}{FB}=\frac{AX}{XB}\).

By Ceva, \(\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right)\frac{AF}{FB}=1\). The first two factors give \(\frac{1}{2}\). Hence \(\frac{AF}{FB}=2\). The positive sign shows that \(F\) lies on segment \(AB\).

Multiply: \(2\cdot\left(-\frac{3}{5}\right)\cdot\frac{5}{6}=-1\). Therefore, by Menelaus' theorem with directed ratios, points \(D,E,F\) lie on one line.

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.