SECTION 1.2 Triangle Geometry 5
Proof. Note first that 4 AA′C′
and 4 CA′C′clearly have the same
areas, which implies that 4 ABC′
and 4 CA′B have the same area
(being the previous common area
plus the area of the common trian-
gle 4 A′BC′). Therefore
A′B
AB
=
1
2 h·A′B
1
2 h·AB
=area 4 A′BC′
area 4 ABC′
=area 4 A′BC′
area 4 CA′B=1
2 h′·BC′
1
2 h′·BC
=
BC′
BC
In an entirely similar fashion one can prove that
A′B
AB
=
A′C′
AC
.
Conversely, assume that
A′B
AB=
BC′
BC
.
In the figure to the right, the point
C′′ has been located so that the seg-
ment [A′C′′] is parallel to [AC]. But
then triangles 4 ABC and 4 A′BC′′
are similar, and so
BC′′
BC=
A′B
AB
=
BC′
BC
,
C"CAC'BA'i.e., thatBC′′=BC′. This clearly implies thatC′=C′′, and so [A′C′]
is parallel to [AC]. From this it immediately follows that triangles