SECTION 1.6 Mass Point Geometry 49
The pointF is located at the center of mass—in particular it is on the
line segments [AD] and [CE]; furthermore its total mass is^172. As a
result, we have thatAF :FD= 7 :^32 = 14 : 3 andCF :FE =^72 : 5 =
14 : 10, in agreement with what was proved above.
We mention in passing that mass point geometry can be used to
prove Ceva’s theorem (and its converse) applied to 4 ABC when the
Cevians [AX], [BY], and [CZ] meet the triangle’s sides [BC], [AC],
and [AB], respectively. If we are given that
AZ
ZB×
BX
XC
×
CY
Y A
= 1,
we assign massZB to vertexA, massAZto vertexB, and massAZXC·BA
to vertexC. Since ZB : AZXC·BX = CYY A, we see that the center of mass
will lie on the intersection of the three Cevians above. Conversely,
if we’re given the three concurrent Cevians [AX], [BY], and [CZ],
then assigning masses as above will place the center of mass at the
intersection of the Cevians [AX] and [CZ]. Since the center of mass is
also on the Cevian [BY], we infer that
CY
Y A=
ZB·XC
AZ·BX
,
and we’re done!
We turn to a few examples, with the hopes of conveying the utility
of this new approach. We emphasize: the problems that follow can
all be solved without mass point geometry; however, the mass point
approach is often simpler and more intuitive!