Similarly,
XC
2245
CY
and
CF
C
is the perpendicular bisector of
XY
,
the third side of
XYZ .
Similarly,
ZB
2245
BX
and
BF
B
is the perpendicular bisector of
ZX
,
a second side of
XYZ .
Since
AF
A
⊥
BC
and
BC
A
,
AF
A
⊥
A
intersecting at
A.
Similarly,
BF
B
⊥
B
intersecting at
B
and
CF
C
⊥
C
intersecting at
C .
A
C
The Orthocenter, Excenters, Excircles, and the Euler Line
(from Sections 4.6 and 4.7)
Therefore, the altitudes of
ABC ,
AF
A
,
BF
B
,
and
CF
C
are concurrent at point H .
Q E D
Therefore, by the "Perpendicular Bisector Concurrence" Theorem, Theorem 4.6.1,
the perpendicular bisectors of the sides of
XYZ ,
AF
A
, BF
B
, and CF
C
are concurrent,
say at point H.
Now,
Quadrilaterals ABCY and AZBG are parallograms.
By the "Opposite Sides of a Parallogram" Theorem,
YA
2245
CB
and
CB
2245
AZ
;
so,
YA
2245
AZ
by transitivity of congruence.
Therefore,
AF
A
is the perpendicular bisector of
YZ
,
one side of
XYZ .
We show that
AF
A
, BF
B
, and CF
C
are the
perpendicular bisectors of the sides of
XYZ ,
which are concurrent by
the "Perpendicular Bisector
Concurrence" Theorem, Theorem 4.6.1.
B
F
C
F
A
F
B
Theorem 4.6.4, The "Altitude Concurrence" Theorem:
The three altitudes of a triangle ABC are concurrent
at a point H .
The point of concurrence
H
is called the Orthocenter of the triangle.
Proof:
Let
ABC be any triangle.
Construct lines
A
,
B
, and
C
,
through A, B, and C and parallel to the side opposite these points, respectively.
Let
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Shirley
 Elementary geometry

Click to edit the document details