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The Orthocenter, Excenters, Excircles, and the Euler Line
(from Sections 4.6 and 4.7)
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 X , Y and Z be the points where
B
and
C
,
A
and
C
, and
A
and
C
intersect in pairs, respectively.
Let FA , FB , and FC be the feet of the altitudes drawn from A, B, and C to the opposite sides (perhaps extended),
respectively.
We show that AFA , BFB , and CFC are the
perpendicular bisectors of the sides of XYZ ,
which are concurrent by the "Perpendicular Bisector
Concurrence" Theorem, Theorem 4.6.1.
Since AFA ⊥ BC and BC
AFA ⊥
B
Y
A
CFC ⊥
B
C
FB
,
FA
H
B
intersecting at B
FC
A
intersecting at C .
A
Z
Now, Quadrilaterals ABCY and AZBG are parallograms.
By the "Opposite Sides of a Parallogram" Theorem,
YA ≅ CB and CB ≅ AZ ;
so,
YA ≅ AZ by transitivity of congruence.
Therefore, AFA is the perpendicular bisector of YZ , one side of
XYZ .
Similarly,
ZB ≅ BX
and
BFB is the perpendicular bisector of ZX , a second side of
XYZ .
Similarly,
XC ≅ CY
and
CFC is the perpendicular bisector of XY , the third side of
XYZ .
Therefore, by the "Perpendicular Bisector Concurrence" Theorem, Theorem 4.6.1,
the perpendicular bisectors of the sides of
XYZ ,
AFA , BFB , and CFC are concurrent,
say at point H.
Therefore, the altitudes of
ABC ,
X
C
intersecting at A.
BFB ⊥
Similarly,
and
A
C
AFA , BFB , and CFC are concurrent at point H .
QED
Theorem 4.6.5, The "Excenter" Theorem:
The angle bisectors of two exterior angles and the angle bisector of the interior angle
opposite their common side are three concurrent rays.
The point of concurrence is called
the "Excenter of the triangle corresponding to the interior angle bisected".
Corollary 4.6.6, The "Excircle" Theorem:
There is a circle centered at each Excenter called
the "Excircle corresponding to the Excenter at its center".
This Excircle is externally tangent to all three sides of the triangle (with two sides extended).
Proof of both: Triangle ABC is given.
Let P be the point of intersection of the bisectors of two external angles of triangle ABC,
say the external angles at vertex B and at vertex C.
Let p 1 be the perp. distance from P to side AB (extended).
Let p 2 be the perp. distance from P to side BC .
Let p 3 be the perp. distance from P to side AC (extended).
By the "Point on the Angle Bisector" Theorem, Theorem 4.6.2, on p. 182,
p 1 = p 2 since P is on the angle bisector of exterior angle B,
and p 2 = p 3 since P is on the angle bisector of exterior angle C.
Since p 1 = p 2 and p 2 = p 3, p 1 = p 3 by transitivity.
By the "Point on the Angle Bisector" Theorem,
Theorem 4.6.2, point P is on the angle bisector of angle ∠ A .
Thus, the three angle bisectors are concurrent at P.
Let F 1 , F 2 and F 3 be the feet of the perpendicular
segments from P to the three sides of the triangle,
AB (extended), BC, and AC (extended),
respectively.
Then, these points F 1 , F 2 and F 3
are all equidistant from point P,
each with the distance p 1 = p 2 = p 3.
So, the circle C( P, p 1 ) centered at P passes
through all three feet of the perpendicular
segments, segments which are radii perpendicular
to the sides (extended) of the triangle.
A
B
C
F2
F1
F3
p2
p1
Thus, by the "Perpendiculars are Tangent" Theorem,
this circle is tangent to the sides of the triangle,
AB (extended), BC, and AC (extended).
p3
P
Since it is in the exterior of the triangle, so it is externally tangent to
the three sides of the triangle (some extended) .
QED
Note: There are three Excenters and three Excircles for each triangle
ABC.
Theorem 4.7.1, The "Altitude Segment" Theorem:
The length of the segment of a triangle's altitude from the vertex to the Orthocenter
is twice the length of the segment from the Circumcenter to the midpoint of the side
opposite that vertex. That is, in the figure below, BH = 2 (SF) .
Proof: Let
ABC be a given triangle.
Let H be its Orthocenter and let S be its Circumcenter.
Extend CS along CS to the point C' on the circumcircle of the triangle.
CAC' and CBC' are semicircles.
Note that AH and BH are the altitudes of the triangle through A and B, resp.
∠CAC' is a right angle since it is inscribed in a semicircle. ∴ AC' ⊥ AC .
Also, since BH lies along an alititude, BH ⊥ AC .
∴ BH
C'A , since both are perpendicular to AC .
∠CBC' is a right angle since it is inscribed in a semicircle. ∴ BC' ⊥ BC .
Also, since AH lies along an alititude, AH ⊥ BC .
∴ AH
B
BC' , since both are perpendicular to BC .
∴ Quadrilateral AHBC' is a parallelogram.
C'
A'
∴ AC' = BH , by the "Opposite Sides
of a Parallelogram" Theorem.
H
Let F be the midpoint of AC .
Circumcenter S is the midpoint of CC' ,
which is a diameter of the circumcircle.
A
By the Midpoint Connection Theorem
applied to
QED
CAC', SF = (1/2) (AC') .
∴ AC' = 2 (SF) .
S
F
C
Theorem 4.7.2, The "Centroid on the Euler Line" Theorem:
The Centroid G is collinear with the Orthocenter H and Circumcenter S ;
Thus, the Centroid G is on the Euler Line.
Also, the Centroid G is located 2 / 3 of the way
along the Euler Line from the Orthocenter to the Circumcenter.
Proof:
Let
ABC be a given triangle. Let points S, H M, F and G be as follows:
S = The Circumcenter
H = The Orthocenter
M = The Midpoint of the side opposite B
F = Foot of altitude from B
G = Centroid
Draw lines HS and BM and altitude BH.
(A technical part of the proof shows that BM and HS intersect.)
Let K be the point where the median BM intersects HS.
B
K is shown to be the Centroid G.
BF || SM since both are ⊥ AC .
The vertical angles at K are congruent and,
by the Converse of the AIA, ∠HBK ≅ ∠SMK.
So,
BHK and
S
K
H
A
MSK are similar by AA(A).
F
By the Altitude Segment Lemma, BH / MS = 2 / 1 .
So, BK / MK = 2 / 1 and HK / SK = 2 / 1 .
So, BK / BM = 2 / 3 and HK / HS = 2 / 3 .
So, K is the point located 2 / 3 of the way along the median BM
from vertex B to midpoint M of the opposite side.
So, by the corollary to the Median Concurrence Theorem,
K = G, the Centroid of ABC .
Since K = G, HG / HS = 2 / 3 .
Thus, the Centroid G of ABC and it is located 2 / 3 of the way
along the Euler Line from the Orthocenter to the Circumcenter.
QED
Euler Line
M
C
K=G