4.1 Geometry 205A M N B
CP
Q
D
Figure 24a problem in plane geometry there is no danger in identifying the areas with the cross-
products, provided that we keep track of the orientation. The hypothesis of the problem
implies that
1
2(
−→
DA×
−−→
DQ+
−−→
AM×
−→
AQ)=
1
2
(
−→
CP×
−→
CB+
−→
BP×
−→
BN).
Hence
−→
DA×
−−→
DQ+
−−→
AM×(
−→
AD+
−−→
DQ)=
−→
CP×
−→
CB+(
−→
BC+
−→
CP)×
−→
BN.
Because
−→
BN=−
−−→
AMand−→
CP=−
−−→
DQ, this equality can be rewritten as(−−→
AM+
−−→
DQ)×(
−→
AD+
−→
CB)= 2
−−→
DQ×
−−→
AM.
Usingthe fact that
−→
AD+
−→
CB=
−→
AB+
−→
CD(which follows from−→
AB+
−→
BC+
−→
CD+
−→
DA=
−→
0 ), we obtain
−−→
AM×−→
CD+
−−→
DQ×
−→
AB= 2
−−→
DQ×
−−→
AM.
From here we deduce that
−−→
AM×
−→
QC=
−−→
DQ×
−−→
MB. These two cross-products point
in opposite directions, so equality can hold only if both are equal to zero, i.e., ifABis
parallel toCD.
Moreapplicationsof the dot and cross-products to geometry can be found below.582.Given two trianglesABCandA′B′C′with the same centroid, prove that one can
construct a triangle with sides equal to the segmentsAA′,BB′, andCC′.
583.Given a quadrilateralABCD, consider the pointsA′,B′,C′,D′on the half-lines
(i.e., rays)|AB,|BC,|CD, and|DA, respectively, such thatAB=BA′,BC=
CB′,CD=DC′,DA=AD′. Suppose now that we start with the quadrilateral
A′B′C′D′. Using a straightedge and a compass only, reconstruct the quadrilateral
ABCD.