# Terquem's Theorem

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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

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Copyright © 1996-2018 Alexander Bogomolny