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