Gergonne and Van Obel theoremsCeva's theorem is a fundamental result in Triangle Geometry that relates ratios of segments cut by the concurrent cevians on the sides of a triangle. A less known result valid in the same configuration has been proved by Van Obel and bears his name. A related theorem discovered by Gergonne is virtually unknown. The latter two have simple proofs based on three observations:
Based on the above, I shall prove the two theorems, Gergonne's coming first: Theorem (Gergonne, 1818)Let three cevians AD, BE, and CF concur at a point K inside ΔABC. Then
and also
ProofSince, for example, AK/AD + KD/AD = 1, (1) and (2) are clearly equivalent. So let's prove (1). By , Area(BKC)/Area(ABC) = KD/AD, Area(AKC)/Area(ABC) = KE/BE, Area(AKB)/Area(ABC) = KE/CE, which, when added up, yield KD/AD + KE/BE + KF/CF = (Area(BKC) + Area(AKC) + Area(AKB))/Area(ABC) = 1. The proof of Van Obel's theorem (often, van Aubel's theorem) takes a little longer. Theorem (Van Obel, van Aubel)AK/KD = AF/FB + AE/EC. ProofBy 2, Area(AKB)/Area(BKD) = AK/KD = Area(AKC)/Area(CKD). By 3,
On the other hand, again by 2, Area(ABE)/Area(BCE) = AE/EC = Area(AKE)/Area(CKE). From here, (Area(ABE) - Area(AKE))/(Area(BCE) - Area(CKE)) = AE/EC. So that
Similarly, (4) and its analogues furnish an extra proof of Ceva's theorem. Indeed
which after multiplication give Ceva's identity. RemarkAs with barycentric coordinates, if the point K lies outside ΔABC, the areas of one or two of the triangles AKB, BKC, AKC may be regarded negative, in which case all three theorems still hold provided the segments involved are also considered signed. |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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