SECTION 1.2 Triangle Geometry 13
1.2.5 Consequences of the Ceva and Menelaus theorems
As one typically learns in an elementary geometry class, there are sev-
eral notions of “center” of a triangle. We shall review them here and
show their relationships to Ceva’s Theorem.
Centroid. In the triangle 4 ABC
lines (AX),(BY), and (CZ)
are drawn so that (AX) bisects
[BC], (BY) bisects [CA], and
(CZ) bisects [AB] That the lines
(AX),(BY), and (CZ) are con-
current immediately follows from
Ceva’s Theorem as one has that
AZ
ZB·
BX
XC
·
CY
Y Z
= 1× 1 ×1 = 1.
The point of concurrency is called thecentroid of 4 ABC. The three
Cevians in this case are calledmedians.
Next, note that if we apply the Menelaus’ theorem to the triangle
4 ACXand the transversal defined by the pointsB, Y and the centroid
P, then we have that
1 =
AY
Y C
·
CB
BX
·
XP
PA
⇒
1 = 1· 2 ·
XP
PA
⇒
XP
PA
=
1
2
.
Therefore, we see that the distance of a triangle’s vertex to the centroid
is exactly 1/3 the length of the corresponding median.