Exercise 1985Geometry › Plane geometryOlympiadAdded 5 Nov 2026

Harmonic division from Ceva and Menelaus

#Ceva#Menelaus#harmonic conjugates#cevians

Statement

Let $D,E,F$ be points on the sides $BC,CA,AB$ of triangle $ABC$ (interior to the sides) such that $AD$ , $BE$ , $CF$ are concurrent. Suppose that line $EF$ is not parallel to $BC$ and meets line $BC$ at $G$ .

Prove that $\dfrac{BD}{DC}=\dfrac{BG}{GC}$ .
Conclude that $G$ lies outside segment $BC$ and that $D$ and $G$ divide $BC$ internally and externally in the same ratio.
Deduce that if $D$ is the midpoint of $BC$ , then $EF\parallel BC$ .

Hint

Write Ceva's theorem for the concurrent lines and Menelaus' theorem for the line $EFG$ ; both contain the same product $\frac{CE}{EA}\cdot\frac{AF}{FB}$ .

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Solution

1. Ceva's theorem (concurrent cevians):
$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1.$
Menelaus' theorem for the triangle $ABC$ and the line $EFG$ (in terms of unsigned lengths):
$\frac{BG}{GC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1.$
Dividing the two relations gives $\dfrac{BD}{DC}=\dfrac{BG}{GC}$ .

2. A line cannot meet all three open sides of a triangle, and $E,F$ are interior to the sides $CA,AB$ ; hence the line $EF$ meets line $BC$ outside the segment $BC$ . So $G$ is outside segment $BC$ while $D$ is inside: $D$ is the internal and $G$ the external point of division in the same ratio, i.e. $(B,C;D,G)=-1$ .

3. If $BD=DC$ then $\frac{CE}{EA}\cdot\frac{AF}{FB}=1$ by Ceva, i.e. $\frac{AF}{FB}=\frac{AE}{EC}$ . By the converse of Thales' theorem $EF\parallel BC$ .

Remark. If $AD$ is the bisector of $\angle BAC$ and $AG$ the external bisector, then $D$ and $G$ are harmonic conjugates.

Lesson. Ceva and Menelaus share a common product; dividing one by the other isolates the harmonic ratio.

Harmonic division from Ceva and Menelaus

Statement

Solution

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