## Menelaus From Ceva

Ceva and Menelaus theorem go naturally together.

### Ceva's Theorem

Three Cevians AD, BE, and CF are concurrent iff

(1) | AF/FB · BD/DC · CE/EA = 1 |

holds.

### Menelaus Theorem

Let three points F, D, and E, lie respectively on the sides AB, BC, and AC of ΔABC. Then the points are collinear iff

(2) | AF/BF · BD/CD · CE/AE = 1 |

The theorems are in fact equivalent. That Ceva's theorem is implied by that of Menelaus has already been established. Below, I prove that the converse is also true: Ceva's theorem implies the theorem of Menelaus. The proof requires introduction of several additional lines:

What if applet does not run? |

Let X, Y, Z be the intersections of BE and AD, AD and CF, and BE and CF, respectively. There are several triangles with three concurrent cevians to which we may apply Ceva's theorem:

Identity | Triangle | Cevians | Intersect At | |
---|---|---|---|---|

(1) | AF/FB · BX/XE · EC/CA = 1 | ABE | BC, AX, EF | D |

(2) | CE/EA · AY/YD · DB/BC = 1 | ACD | AB, CY, DE | F |

(3) | BD/DC · CZ/ZF · FA/AB = 1 | BCF | AC, BZ, DF | E |

(4) | EX/XB · BC/CD · DF/FE = 1 | BDE | BF, CE, DX | A |

(5) | DY/YA · AB/BF · FE/ED = 1 | ADF | AE, BD, FY | C |

(6) | FZ/ZC · CA/AE · ED/DF = 1 | CEF | CD, AF, EZ | B |

Now multiply (1)-(6), taking into account the sign convention for directed segments. The product reduces to

(7) | (AF/BF · BD/CD · CE/AE)^{2} = 1. |

But the straight line DEF is a transversal of ΔABC. It could be argued that the points D, E, F either all lie on the extensions of the sides of ΔABC, or only one does, while the other two lie on its sides. A simple verification of signs then shows that the product in (7) is positive:

(8) | AF/BF · BD/CD · CE/AE = 1, |

which proves Menelaus' theorem.

There are two additional proofs of the equivalency of the two theorems. One is derived from a solution of the 4-travelers problem and the other from the observation of poles and polars with respect to a triangle and the Desargues' theorem.

### References

- C. W. Dodge,
*Euclidean Geometry and Transformations*, Dover, 2004 (reprint of 1972 edition), pp. 22-23. - J. R. Silvester,
*Ceva = (Menelaus)*,^{2}__The Math. Gazette__, v 84, N 5 (2000), pp 268-271.

### Desargues' Theorem

- Desargues' Theorem
- 2N-Wing Butterfly Problem
- Cevian Triangle
- Do You Speak Mathematics?
- Desargues in the Bride's Chair (with Pythagoras)
- Menelaus From Ceva
- Monge from Desargues
- Monge via Desargues
- Nobbs' Points, Gergonne Line
- Soddy Circles and David Eppstein's Centers
- Pascal Lines: Steiner and Kirkman Theorems II
- Pole and Polar with Respect to a Triangle
- Desargues' Hexagon
- The Lepidoptera of the Circles

### Menelaus and Ceva

- The Menelaus Theorem
- Menelaus Theorem: proofs ugly and elegant - A. Einstein's view
- Ceva's Theorem
- Ceva in Circumscribed Quadrilateral
- Ceva's Theorem: A Matter of Appreciation
- Ceva and Menelaus Meet on the Roads
- Menelaus From Ceva
- Menelaus and Ceva Theorems
- Ceva and Menelaus Theorems for Angle Bisectors
- Ceva's Theorem: Proof Without Words
- Cevian Cradle
- Cevian Cradle II
- Cevian Nest
- Cevian Triangle
- An Application of Ceva's Theorem
- Trigonometric Form of Ceva's Theorem
- Two Proofs of Menelaus Theorem
- Simultaneous Generalization of the Theorems of Ceva and Menelaus
- Menelaus from 3D
- Terquem's Theorem
- Cross Points in a Polygon
- Two Cevians and Proportions in a Triangle, II
- Concurrence Not from School Geometry
- Two Triangles Inscribed in a Conic - with Elementary Solution
- From One Collinearity to Another
- Concurrence in Right Triangle
|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

66163934