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


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Discussion

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

The 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.)


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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

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

Menelaus and Ceva

 61228369

Search by google: