SECTION 1.2 Triangle Geometry 9
Lemma. Given the triangle
4 ABC, letXbe the intersection of
a line throughAand meeting(BC).
LetP be any other point on(AX).
Then
area 4 APB
area 4 APC=
BX
CX
.
Proof. In the diagram to the
right, altitudes BR and CS have
been constructed. From this, we see
that
area 4 APB
area 4 APC=
1
2 AP·BR
1
2 AP·CS
=BR
CS
=
BX
CX
,
where the last equality follows from the obvious similarity
4 BRX∼4CSX.
Note that the above proof doesn’t depend on where the line (AP) in-
tersects (BC), nor does it depend on the position ofP relative to the
line (BC), i.e., it can be on either side.
Ceva’s Theorem. Given the triangle 4 ABC, lines (usually called
Ceviansare drawn from the verticesA,B, andC, withX,Y, andZ,
being the points of intersections with the lines(BC), (AC), and(AB),
respectively. Then(AX), (BY),and(CZ)are concurrent if and only
if