14 CHAPTER 1 Advanced Euclidean Geometry
Orthocenter. In the trian-
gle 4 ABC lines (AX),(BY),and
(CZ) are drawn so that (AX) ⊥
(BC), (BY)⊥(CA), and (CZ) ⊥
(AB). Clearly we either have
AZ
ZB,
BX
XC
,
CY
Y A
> 0
or that exactly one of these ratios
is positive. We have
4 ABY ∼4ACZ ⇒AZ
AY
=
CZ
BY
.
Likewise, we have4 ABX∼4CBZ ⇒BX
BZ
=
AX
CZ
and 4 CBY ∼4CAX⇒
CY
CX
=
BY
AX
.
Therefore,
AZ
ZB·
BX
XC
·
CY
Y A
=
AZ
AY
·
BX
BZ
·
CY
CX
=
CZ
BY
·
AX
CZ
·
BY
AX
= 1.
By Ceva’s theorem the lines (AX),(BY),and (CZ) are concurrent, and
the point of concurrency is called theorthocenter of 4 ABC. (The
line segments [AX],[BY], and [CZ] are thealtitudesof 4 ABC.)
Incenter. In the triangle 4 ABC lines
(AX),(BY), and (CZ) are drawn so
that (AX) bisects BAĈ , (BY) bisects
ABĈ , and (CZ) bisects BCÂ As we
show below, that the lines (AX),(BY),
and (CZ) are concurrent; the point of
concurrency is called the incenter of
4 ABC. (A very interesting “extremal”