# Terquem's Theorem

What Is This About?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander BogomolnyThe applet is supposed to suggest a theorem attributed to the French-Jewish mathematican Olry Terquem (1782-1862):

Let in ΔABC the cevians through point M meet the sides AB, BC, and AC (or their extensions) in points D, E, F, respectively. The circumcircle C(DEF) of ΔDEF, crosses the sides of ΔABC in 3 other points, D', E', F'. Then the cevians AD', BE', CF' are concurrent (point M' in the applet.)

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### Proof

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

- 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 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
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Copyright © 1996-2018 Alexander Bogomolny

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