Cevians, Symmedians, and
Excircles
MA 341 – Topics in Geometry
Lecture 16
Cevian
A cevian is a line segment which joins a vertex of a
triangle
l with
h a point on the
h opposite side
d (or
( its
extension).
B
cevian
A
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Cevian Triangle & Circle
• Pick P in the interior of ∆ABC
• Draw cevians from
f
each
h vertex through
h
h P to the
h
opposite side
• Gives set of three intersecting cevians AA
AA’, BB
BB’,
and CC’ with respect to that point.
• The triangle ∆A
∆A’B’C’
B C is known as the cevian
triangle of ∆ABC with respect to P
• Circumcircle of ∆A’B’C’ is known as the evian
circle with respect to P.
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Cevian
circle
Cevian
C
i
triangle
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Cevians
In ∆ABC examples of cevians are:
medians
di
– cevian
i point
i =G
perpendicular bisectors – cevian point = O
angle bisectors – cevian point = I (incenter)
altitudes – cevian point = H
Ceva’s
’ Theorem
h
deals
d l with
h concurrence of
f any
set of cevians.
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Gergonne Point
In ∆ABC find the incircle and points of
tangency of
f incircle
i i l with
i h sides
id of
f ∆ABC.
∆ BC
Known as contact triangle
g
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Gergonne Point
These cevians are concurrent!
Why?
Recall that AE=AF, BD=BF, and CD=CE
Ge
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Gergonne Point
The point is called the Gergonne point, Ge.
Ge
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Gergonne Point
Draw lines parallel to sides of contact
g through
ug Ge..
triangle
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Gergonne Point
Six points are concyclic!!
Called the Adams Circle
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Gergonne Point
Center of Adams circle = incenter of ∆ABC
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Isogonal Conjugates
Two lines AB and AC through vertex A are
g
if one is the reflection
said to be isogonal
of the other through the angle bisector.
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Isogonal Conjugates
If lines through A, B, and C are concurrent
g
lines are concurrent
at P,, then the isogonal
at Q.
Points P and Q are isogonal conjugates.
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Symmedians
In ∆ABC, the symmedian ASa is a cevian
g vertex A ((Sa  BC)) isogonally
g
y
through
conjugate to the median AMa, Ma being the
p
of BC.
midpoint
The other two
symmedians BSb
and CSc are defined
y
similarly.
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Symmedians
The three symmedians ASa, BSb and CSc
concur in a p
point commonly
y denoted K and
variably known as either
• the symmedian point or
• the Lemoine point
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Symmedian of Right Triangle
The symmedian point K of a right triangle
is the midpoint of the altitude to the
hypotenuse A
hypotenuse.
Mb
K
B
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Proportions of the Symmedian
Draw the cevian from vertex A,
A through
the symmedian point, to the opposite side
of the triangle,
triangle meeting BC at Sa. Then
c
b
BSa c2
 2
CSa b
a
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Length of the Symmedian
Draw the cevian from vertex C,
C through
the symmedian point, to the opposite side
of the triangle.
triangle Then this segment has
length
2
2
2
ab 2a  2b  c
CSc 
a2  b2
2
2
2
Likewise
bc 2b  2c  a
ASa 
2
2
b c
ac 2a  2c  b
B b
BS
2
2
a c
2
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Excircles
In several versions of geometry triangles
are defined in terms of lines not
segments.
A
C
B
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Excircles
Do these sets of three lines define
circles?
Known as tritangent circles
A
C
B
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Excircles
A
IC
rc
IB
rb
I
B
C
IA
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ra
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Construction of Excircles
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Extend the sides
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Bisect exterior angle at A
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Bisect exterior angle at B
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Find intersection
Ic
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Drop perpendicular to AB
Ic
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Find point of intersection with
AB
Ic
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Construct circle centered at Ic
Ic
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rc
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Excircles
The Ia, Ib, and Ic are called excenters.
ra, rb, rc are called
ll d exradii
d
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Excircles
Theorem: The length of the tangent from
a vertex to the opposite exscribed circle
equals the semiperimeter, s.
CP = s
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Excircles
1. CQ
Q = CP
2. AP = AY
3. CP = CA+AP
= CA+AY
Q BC+BY
4. CQ=
5. CP + CQ
Q = AC + AY + BY + BC
6. 2CP = AB + BC + AC = 2s
7. CP = s
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Exradii
1. CPICP
2 tan(C/2)=r
2.
t (C/2) C/s
/
3. Use Law of
Tangents
Ic
C
(s  a)(s  b)
rc  s tan    s

s(s
(  c))
2
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s(s  a)(s  b)
sc
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Exradii
Likewise
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ra 
s(s  b)(s  c)
sa
rb 
s(s  a)(s  c)
s b
rc 
s(s  a)(s  b)
sc
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Excircles
Theorem: For any triangle ∆ABC
1 1
1
1



r r a rb rc
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Excircles
1 1 1
  
ra rb rc

sa
s(s  a)(s  b)(s  c)
s(s  a)(s  b)(s  c)

sc
s(s  a)(s  b)(s  c)
(  a)(s
)(  b)(s
)(  c))
s(s
s

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s b

3s  (a  b  c)


sa
s b
sc


s(s
(  b)(s
)(  c))
s(s
(  a)(
a)(s  c))
s(s
(  a)(
a)(s  b))
s(s  a)(s  b)(s  c)

s
K
1
r
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Nagel Point
In ∆ABC find the excircles and points of
tangency of
f the
h excircles
i l with
i h sides
id of
f
∆ABC.
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Nagel Point
These cevians are concurrent!
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Nagel Point
Point is known as the Nagel point
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Mittenpunkt Point
The mittenpunkt of ∆ABC is the
ymm
point of
p
f the excentral triangle
g
symmedian
(∆IaIbIc formed from centers of
excircles))
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Mittenpunkt Point
The mittenpunkt of ∆ABC is the point of
f the lines f
from
m the
intersection of
excenters through midpoints of
p
g sides
corresponding
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Spieker Point
The Spieker center is center of Spieker
. ., the incenter of
f the medial
m
circle,, i.e.,
triangle of the original triangle .
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Special Segments
Gergonne point, centroid and mittenpunkt
are collinear
GGe
=2
GM
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Special Segments
Mittenpunkt, Spieker center and
orthocenter are collinear
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Special Segments
Mittenpunkt, incenter and symmedian point
K are collinear with
w
distance ratio
IM 2(a2 +b2 + c2 )
=
2
MK
(a +b + c)
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Nagel Line
The Nagel line is the line on which the
g centroid , Spieker
p
incenter , triangle
center Sp, and Nagel point Na lie.
GNa
=2
IG
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Various Centers
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