ELEMENTARYGEOMETRY Search
Menu
Classical

The Centroid through Ceva and Vectors

Two proofs and the further reach of each method

Ceva’s theorem proves the concurrence of the medians and leads naturally to isotomic conjugation. Vectors locate the centroid and reveal why another triangle can share it.

Methods Ceva's theorem · Vectors · Affine geometry · Isotomic conjugation

The theorem

Theorem

Existence of the centroid

Let L,M,NL,M,N be the midpoints of BC,CA,ABBC,CA,AB, respectively. Then the medians AL,BM,CNAL,BM,CN are concurrent at a point GG, called the centroid of ABCABC.

A scalene triangle ABC with L, M, and N at the side midpoints. The medians AL, BM, and CN meet at G.
The three medians of triangle ABC meet at G.

First proof: Ceva’s theorem

Because L,M,NL,M,N are the midpoints of the three sides,

BLLC=1,CMMA=1,ANNB=1.\frac{BL}{LC}=1, \qquad \frac{CM}{MA}=1, \qquad \frac{AN}{NB}=1.

Therefore,

BLLC⋅CMMA⋅ANNB=1.\frac{BL}{LC} \cdot \frac{CM}{MA} \cdot \frac{AN}{NB} =1.

By the converse of Ceva’s theorem,

AL,BM,CNAL,\qquad BM,\qquad CN

are concurrent.

A further use of Ceva: isotomic conjugation

Before turning to vectors, the same product has a useful extension.

Let AD,BE,CFAD,BE,CF be cevians concurrent at an interior point PP, where

D∈BC,E∈CA,F∈AB.D\in BC, \qquad E\in CA, \qquad F\in AB.

By Ceva’s theorem,

BDDC⋅CEEA⋅AFFB=1.\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} =1.

Define D′∈BCD'\in BC, E′∈CAE'\in CA, and F′∈ABF'\in AB by reversing the three internal division ratios:

BD′D′C=DCBD,CE′E′A=EACE,AF′F′B=FBAF.\frac{BD'}{D'C}=\frac{DC}{BD}, \qquad \frac{CE'}{E'A}=\frac{EA}{CE}, \qquad \frac{AF'}{F'B}=\frac{FB}{AF}.

Equivalently, D′D', E′E', and F′F' are the reflections of DD, EE, and FF in the midpoints of the corresponding sides.

Consequently,

BD′D′C⋅CE′E′A⋅AF′F′B=1.\frac{BD'}{D'C} \cdot \frac{CE'}{E'A} \cdot \frac{AF'}{F'B} =1.

The converse of Ceva’s theorem now shows that AD′,BE′,CF′AD',BE',CF' are concurrent. Their common point QQ is the isotomic conjugate of PP.

For the centroid, all three ratios are 1:11:1. Reversing them leaves each cevian foot unchanged, so the corresponding cevians are again the medians. The centroid is fixed by isotomic conjugation.

A scalene triangle ABC with cevian feet D, E, and F defined by internal division ratios. Reversing those ratios gives D-prime, E-prime, and F-prime, whose cevians meet at Q.
Ceva’s theorem beyond the centroid: reversing the three internal division ratios produces Q, the isotomic conjugate of P.

Second proof: vectors

Choose an arbitrary origin OO, and write

OA→=a,OB→=b,OC→=c.\overrightarrow{OA}=\mathbf a, \qquad \overrightarrow{OB}=\mathbf b, \qquad \overrightarrow{OC}=\mathbf c.

Define GG by

OG→=g=a+b+c3=13a+23(b+c2).\begin{aligned} \overrightarrow{OG}=\mathbf g &=\frac{\mathbf a+\mathbf b+\mathbf c}{3}\\ &=\frac13\mathbf a +\frac23\left(\frac{\mathbf b+\mathbf c}{2}\right). \end{aligned}

Since (b+c)/2(\mathbf b+\mathbf c)/2 is the position vector of the midpoint LL of BCBC, this shows that G∈ALG\in AL and GL:GA=1:2GL:GA=1:2. Applying the same calculation cyclically places GG on BMBM and CNCN. Hence the three medians are concurrent at GG, with

G∈AL∩BM∩CN,GL:GA=GM:GB=GN:GC=1:2.G\in AL\cap BM\cap CN, \qquad GL:GA=GM:GB=GN:GC=1:2.

Viewed in terms of the three vertices, the same formula identifies GG as their equal-weight affine average.

A further use of vectors: a shared centroid

Proposition

Let D∈BCD\in BC, E∈CAE\in CA, and F∈ABF\in AB satisfy

BD:DC=CE:EA=AF:FB.BD:DC=CE:EA=AF:FB.

Then the triangles ABCABC and DEFDEF have the same centroid.

Write the common ratio as r:sr:s, where r,s>0r,s>0. Their position vectors are

d=sb+rcr+s,e=sc+rar+s,f=sa+rbr+s.\mathbf d=\frac{s\mathbf b+r\mathbf c}{r+s}, \qquad \mathbf e=\frac{s\mathbf c+r\mathbf a}{r+s}, \qquad \mathbf f=\frac{s\mathbf a+r\mathbf b}{r+s}.

Therefore,

d+e+f3=a+b+c3=g.\frac{\mathbf d+\mathbf e+\mathbf f}{3} =\frac{\mathbf a+\mathbf b+\mathbf c}{3} =\mathbf g.

This proves the proposition.