10/5/2011

1

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

05-Oct-2011 MA 341 001 2

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

05-Oct-2011 MA 341 001 3

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.

10/5/2011

2

Ceviancircle

05-Oct-2011 MA 341 001 4

Ceviantriangle

CeviansIn ∆ABC examples of cevians are:

medians – cevian point = Gperpendicular bisectors – cevian point = Oangle bisectors – cevian point = I (incenter)altitudes – cevian point = H

05-Oct-2011 MA 341 001 5

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

05-Oct-2011 MA 341 001 6

10/5/2011

3

Gergonne PointThese cevians are concurrent! Why?Recall that AE=AF, BD=BF, and CD=CE

05-Oct-2011 MA 341 001 7

Ge

Gergonne PointThe point is called the Gergonne point, Ge.

05-Oct-2011 MA 341 001 8

Ge

Gergonne PointDraw lines parallel to sides of contact triangle through Ge.

05-Oct-2011 MA 341 001 9

10/5/2011

4

Gergonne PointSix points are concyclic!!Called the Adams Circle

05-Oct-2011 MA 341 001 10

Gergonne PointCenter of Adams circle = incenter of ∆ABC

05-Oct-2011 MA 341 001 11

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.

05-Oct-2011 MA 341 001 12

10/5/2011

5

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

05-Oct-2011 MA 341 001 13

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

05-Oct-2011 MA 341 001 14

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

05-Oct-2011 MA 341 001 15

• the Lemoine point

10/5/2011

6

Symmedian of Right Triangle

The symmedian point K of a right triangle is the midpoint of the altitude to the hypotenuse. A

05-Oct-2011 MA 341 001 16

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

05-Oct-2011 MA 341 001 17

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

05-Oct-2011 MA 341 001 18

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

10/5/2011

7

ExcirclesIn several versions of geometry triangles are defined in terms of lines not segments.

05-Oct-2011 MA 341 001 19

A

B C

ExcirclesDo these sets of three lines define circles?Known as tritangent circles

05-Oct-2011 MA 341 001 20

A

B C

Excircles

AIC IB

Irc rb

05-Oct-2011 MA 341 001 21

BC

I

IA

c rb

ra

10/5/2011

8

Construction of Excircles

05-Oct-2011 MA 341 001 22

Extend the sides

05-Oct-2011 MA 341 001 23

Bisect exterior angle at A

05-Oct-2011 MA 341 001 24

10/5/2011

9

Bisect exterior angle at B

05-Oct-2011 MA 341 001 25

Find intersection

Ic

05-Oct-2011 MA 341 001 26

Drop perpendicular to AB

Ic

05-Oct-2011 MA 341 001 27

10/5/2011

10

Find point of intersection with AB

Ic

05-Oct-2011 MA 341 001 28

Construct circle centered at Ic

Ic rc

05-Oct-2011 MA 341 001 29

05-Oct-2011 MA 341 001 30

10/5/2011

11

ExcirclesThe Ia, Ib, and Ic are called excenters.ra, rb, rc are called exradii

05-Oct-2011 MA 341 001 31

ExcirclesTheorem: The length of the tangent from a vertex to the opposite exscribed circle equals the semiperimeter, s.

05-Oct-2011 MA 341 001 32

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

05-Oct-2011 MA 341 001 33

10/5/2011

12

Exradii1. CPICP2. tan(C/2)=rC/s3. Use Law of

TangentsIc

05-Oct-2011 MA 341 001 34

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)

05-Oct-2011 MA 341 001 35

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

05-Oct-2011 MA 341 001 36

10/5/2011

13

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)

05-Oct-2011 MA 341 001 37

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.

05-Oct-2011 MA 341 001 38

Nagel PointThese cevians are concurrent!

05-Oct-2011 MA 341 001 39

10/5/2011

14

Nagel PointPoint is known as the Nagel point

05-Oct-2011 MA 341 001 40

Mittenpunkt PointThe mittenpunkt of ∆ABC is the symmedian point of the excentral triangle (∆IaIbIc formed from centers of excircles)

05-Oct-2011 MA 341 001 41

Mittenpunkt PointThe mittenpunkt of ∆ABC is the point of intersection of the lines from the excenters through midpoints of corresponding sides

05-Oct-2011 MA 341 001 42

10/5/2011

15

Spieker PointThe Spieker center is center of Spiekercircle, i.e., the incenter of the medial triangle of the original triangle .

05-Oct-2011 MA 341 001 43

Special SegmentsGergonne point, centroid and mittenpunktare collinear

GGe =2

05-Oct-2011 MA 341 001 44

=2GM

Special SegmentsMittenpunkt, Spieker center and orthocenter are collinear

05-Oct-2011 MA 341 001 45

10/5/2011

16

Special SegmentsMittenpunkt, incenter and symmedian point K are collinear with distance ratio

2 2 2

2IM 2(a +b +c )=MK (a +b+c)

05-Oct-2011 MA 341 001 46

Nagel LineThe Nagel line is the line on which the incenter , triangle centroid , Spieker center Sp, and Nagel point Na lie.

GNa 2

05-Oct-2011 MA 341 001 47

GNa =2IG

Various Centers

05-Oct-2011 MA 341 001 48