cevians, symmedians, and excircles cevian … · cevians, symmedians, and excircles ma 341 –...
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Cevians, Symmedians, and Excircles
MA 341 – Topics in GeometryLecture 16
CevianA cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension).
B
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A CD
cevian
Cevian Triangle & Circle• Pick P in the interior of ∆ABC• Draw cevians from each vertex through P to the
opposite side • Gives set of three intersecting cevians AA’, BB’,
and CC’ with respect to that point
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and CC with respect to that point. • The triangle ∆A’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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Ceviancircle
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Ceviantriangle
CeviansIn ∆ABC examples of cevians are:
medians – cevian point = Gperpendicular bisectors – cevian point = Oangle bisectors – cevian point = I (incenter)altitudes – cevian point = H
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altitudes cevian point = H
Ceva’s Theorem deals with concurrence of any set of cevians.
Gergonne PointIn ∆ABC find the incircle and points of tangency of incircle with sides of ∆ABC. Known as contact triangle
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Gergonne PointThese cevians are concurrent! Why?Recall that AE=AF, BD=BF, and CD=CE
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Ge
Gergonne PointThe point is called the Gergonne point, Ge.
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Ge
Gergonne PointDraw lines parallel to sides of contact triangle through Ge.
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Gergonne PointSix points are concyclic!!Called the Adams Circle
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Gergonne PointCenter of Adams circle = incenter of ∆ABC
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Isogonal ConjugatesTwo lines AB and AC through vertex A are said to be isogonal if one is the reflection of the other through the angle bisector.
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Isogonal ConjugatesIf lines through A, B, and C are concurrent at P, then the isogonal lines are concurrent at Q.
Points P and Q are isogonal conjugates
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Points P and Q are isogonal conjugates.
SymmediansIn ∆ABC, the symmedian ASa is a cevianthrough vertex A (Sa BC) isogonallyconjugate to the median AMa, Ma being the midpoint of BC. The other two
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The other two symmedians BSband CSc are defined similarly.
SymmediansThe three symmedians ASa, BSb and CScconcur in a point commonly denoted K and variably known as either• the symmedian point or• the Lemoine point
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• 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
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B CD
MbK
Proportions of the Symmedian
Draw the cevian from vertex A, through the symmedian point, to the opposite side of the triangle, meeting BC at Sa. Then
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2
a2
a
BS cCS b
c
b
a
Length of the Symmedian
Draw the cevian from vertex C, through the symmedian point, to the opposite side of the triangle. Then this segment has length 2 2 2b 2 2blength
Likewise
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2 2 2
c 2 2ab 2a 2b cCS
a b
2 2 2
a 2 2
2 2 2
b 2 2
bc 2b 2c aASb c
ac 2a 2c bBSa c
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ExcirclesIn several versions of geometry triangles are defined in terms of lines not segments.
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A
B C
ExcirclesDo these sets of three lines define circles?Known as tritangent circles
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A
B C
Excircles
AIC IB
Irc rb
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BC
I
IA
c rb
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 rc
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ExcirclesThe Ia, Ib, and Ic are called excenters.ra, rb, rc are called exradii
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ExcirclesTheorem: The length of the tangent from a vertex to the opposite exscribed circle equals the semiperimeter, s.
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CP = s
1. CQ = CP2. AP = AY3. CP = CA+AP
= CA+AY
Excircles
4. CQ= BC+BY
5. CP + CQ = AC + AY + BY + BC6. 2CP = AB + BC + AC = 2s7. CP = s
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Exradii1. CPICP2. tan(C/2)=rC/s3. Use Law of
TangentsIc
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cC (s a)(s b) s(s a)(s b)r stan s2 s(s c) s c
ExradiiLikewise
as(s b)(s c)r
s as(s a)(s c)
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b
c
s(s a)(s c)rs b
s(s a)(s b)rs c
ExcirclesTheorem: For any triangle ∆ABC
a b c
1 1 1 1r r r r
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Excircles
a b c
1 1 1 s a s b s cr r r s(s b)(s c) s(s a)(s c) s(s a)(s b)
s a s b s cs(s a)(s b)(s c) s(s a)(s b)(s c) s(s a)(s b)(s c)
3s (a b c)
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3s (a b c)s(s a)(s b)(s c)
s sKs(s a)(s b)(s c)
1r
Nagel PointIn ∆ABC find the excircles and points of tangency of the excircles with sides of ∆ABC.
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Nagel PointThese cevians are concurrent!
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Nagel PointPoint is known as the Nagel point
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Mittenpunkt PointThe mittenpunkt of ∆ABC is the symmedian point of the excentral triangle (∆IaIbIc formed from centers of excircles)
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Mittenpunkt PointThe mittenpunkt of ∆ABC is the point of intersection of the lines from the excenters through midpoints of corresponding sides
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Spieker PointThe Spieker center is center of Spiekercircle, i.e., the incenter of the medial triangle of the original triangle .
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Special SegmentsGergonne point, centroid and mittenpunktare collinear
GGe =2
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=2GM
Special SegmentsMittenpunkt, Spieker center and orthocenter are collinear
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Special SegmentsMittenpunkt, incenter and symmedian point K are collinear with distance ratio
2 2 2
2IM 2(a +b +c )=MK (a +b+c)
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Nagel LineThe Nagel line is the line on which the incenter , triangle centroid , Spieker center Sp, and Nagel point Na lie.
GNa 2
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GNa =2IG
Various Centers
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