Terquem's Theorem
What Is This About?
A Mathematical Droodle

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Discussion

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According to Ceva's theorem,

AF/BF · BD/CD · CE/AE = 1.

By the Power of a Point theorem,

AF·AF' = AE·AE',
BD·BD' = BF·BF',
CE·CE' = CD·CD'.

The four together reduce to

AF'/BF' · BD'/CD' · CE'/AE' = 1

which, by the converse of Ceva's theorem, implies the required concurrency of AD', BE', and CF'.

References

1. F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, p. 558

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 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
  • Theorems of Ceva and Menelaus, an Illustrated Generalization
• Menelaus from 3D
• Terquem's Theorem
• Cross Points in a Polygon
• Two Cevians and Proportions in a Triangle, II
  

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  Copyright © 1996-2012 Alexander Bogomolny