Advanced Euclidean Geometry
Paul Yiu
Department of MathematicsFlorida Atlantic University
Summer 2016
July 11
Menelaus and Ceva Theorems
Menelaus’ theorem
Theorem 0.1 (Menelaus). Given a triangle ABC with points X , Y , Z onthe side lines BC, CA, AB respectively, the points X , Y , Z are collinear ifand only if
BX
XC· CY
Y A· AZZB
= −1.
A
B CX
Y
Z
W
Yiu: Advanced Euclidean Geometry 2016
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Menelaus’ theorem
Proof. (=⇒)
A
B CX
Y
Z
W
Let W be the point on AC such that BW//XY . Then,
BX
XC=
WY
Y C, and
AZ
ZB=
AY
YW.
It follows that
BX
XC· CY
Y A· AZZB
=WY
Y C· CY
Y A· AYYW
=CY
Y C· AYY A
· WY
YW= −1.
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Menelaus theorem
(⇐=) Suppose the line joining X and Z intersects AC at Y ′. From above,
BX
XC· CY ′
Y ′A· AZZB
= −1 =BX
XC· CY
Y A· AZZB
.
It follows thatCY ′
Y ′A=
CY
Y A.
The points Y ′ and Y divide the segment CA in the same ratio. These mustbe the same point, and X , Y , Z are collinear.
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Example: The external bisectors
The external angle bisectors of a triangle intersect their opposite sides atthree collinear points.
a
bc
B
C
A
Y ′
Z′
X′
Proof. If the external bisectors are AX ′, BY ′, CZ ′ with X ′, Y ′, Z ′ on BC,CA, AB respectively, then
BX ′
X ′C= −c
b,
CY ′
Y ′A= −a
c,
AZ ′
Z ′B= − b
a.
It follows that BX ′X ′C · CY ′
Y ′A · AZ ′Z ′B = −1 and the points X ′, Y ′, Z ′ are collinear.
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Example
Given triangle ABC and pointsY on CA, Z on AB, and X on the extension of BC,such that X , Y , Z are collinear.If CX = x, BZ = z, and AY = y,and two of these lengths are given, calculate the remaining one.
c
ba
x
y
z AB
C
Z
X
Y
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c
ba
x
y
z AB
C
Z
X
Y
(a, b, c) x y z
(3, 4, 5) 1 2 4
(3, 5, 6) 1 4
(3, 5, 7) 1 6
(4, 5, 6) 3 4
(4, 5, 7) 2 4
(5, 6, 7) 1 6
(6, 7, 8) 4 6
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(2, 3, 4) triangle
ABC is a triangle with a = 2, b = 3, c = 4.A transversal intersects the sidelines at X , Y , Zsuch that AY = BZ = CX = t.Calculate t.
4
32
AB
C
Z
X
Y
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(7, 12, 18) triangle
ABC is a triangle with a = 7, b = 12, c = 18.A transversal intersects the sidelines at X , Y , Zsuch that AZ = BX = CY = t.Calculate t.
18
127
AB
C
Z
Y
X
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(9, 10, 12) triangle
ABC is a triangle with a = 9, b = 10, c = 12.A transversal intersects the sidelines at X , Y , Zsuch that BX = CY = AZ = t.Calculate t.
12
109
AB
C
Z
X
Y
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Example
Given three circles with centers A, B, C and distinct radii,show that the exsimilicenters of the three pairs of circles are collinear.
A
B
C
X
Y
Z
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Line with equal intercepts on sidelines of a given triangle
Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internally so that CX = CY = AZ.
x
x
x
A
B C
Z
X
Y
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Solution
Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internally so that CX = CY = AZ.
rr
x
I
A
B C
Z
X
Y
If CX = x, by Menelaus’ theorem, we requirea+ x
−x· x
b− x· x
c− x= −1.
From this,
x =bc
a+ b+ c.
Note that
x =bc sinA
2s sinA=
Δ
s· 1
sinA=
r
sinA.
This means that IZ is parallel to CA, and suggests the following simpleconstruction of the line.
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Construction
r
r
x
I
A
B C
Z
X
Y
(1) Construct the incenter I of triangleABC.(2) Construct a line through I parallel to AC, to intersect AB at Z.(3) Construct a circle with center C, radius AZ, to intersect BC externallyat X and CA internally at Y .
Then X , Y , Z are collinear with CX = CY = AZ = rsinA .
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Line with equal intercepts on sidelines of a given triangle
Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internallyso that CX = CY = BZ.
y
y
y
A
B C
Z
X
Y
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Solution
Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internally so that CX = CY = BZ.
y y
y
A
B C
Z
X
Y
Ac
Bc
Cc
Ic
If CX = CY = BZ = y, by Menelaus’ theorem, we require
a+ y
−y· y
b− y· c− y
y= −1.
From this,y =
ca
a+ b− c=
rcsinB
,
where rc is the radius of the excircle on the sideAB.
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Ceva’s theorem
Theorem 0.2 (Ceva). Given a triangleABC with pointsX , Y , Z on the sidelines BC, CA, AB respectively, the lines AX , BY , CZ are concurrent ifand only if
BX
XC· CY
Y A· AZZB
= +1.
P
X
Y
Z
A
B C
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Proof
P
X
Y
Z
A
B C
Proof. (=⇒) Suppose the lines AX , BY , CZ intersect at a point P . Con-sider the line BPY cutting the sides of triangle CAX . By Menelaus’ theo-rem,
CY
Y A· APPX
· XB
BC= −1, or
CY
Y A· PA
XP· BX
BC= +1.
Also, consider the line CPZ cutting the sides of triangle ABX . ByMenelaus’ theorem again,
AZ
ZB· BC
CX· XP
PA= −1, or
AZ
ZB· BC
XC· XP
PA= +1.
Multiplying the two equations together, we have
CY
Y A· AZZB
· BX
XC= +1.
(⇐=) Exercise.
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Example
Given triangle ABC and points X on BC, Y on CA, Z on AB
such that the cevians AX , BY , CZ are concurrent.If CX = x, BZ = z, and AY = y,and two of these lengths are given, calculate the remaining one.
c
ba
z
y
x
AB
C
Z
X Y
(a, b, c) x y z
(3, 4, 6) 1 2 4
(3, 5, 6) 1 4
(3, 5, 7) 1 3
(4, 5, 6) 3 4
(4, 5, 7) 1 4
(5, 6, 7) 3 4
(6, 7, 9) 5 4
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(3, 4, 6) triangle
ABC is a triangle with a = 3, b = 4, c = 6.X , Y , Z are points on BC, CA, AB respectivelysuch that BZ = AY = CX = t.If the cevians AX , BY , CZ are concurrent, calculate t.
6
43
AB
C
Z
X Y
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Example
ABC is a right triangle.Show that the lines AX , BY , and CQ are concurrent.
Q
PA B
C
Z Z′
Y
Y ′
X′
X
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Solution
ABC is a right triangle.Show that the lines AX , BY , and CQ are concurrent.
Q
Z0A B
C
Z Z′
Y
Y ′X′
X
X0Y0
Let AX intersect BC at X0,BY intersect CA at Y0, and CQ intersect AB at Z0.
AZ0
Z0B· BX0
X0C· CY0
Y0A=
AZ0 · AZZ0B · BZ ′ ·
BX
AC· BC
AY
=AC2
BC2· BC
AC· BC
AC= 1.
By Ceva’s theorem, the lines AX0, BY0, CZ0 are concurrent.Yiu: Advanced Euclidean Geometry 2016
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The centroid
If D, E, F are the midpoints of the sides BC, CA, AB of triangle ABC,then clearly
AF
FB· BD
DC· CE
EA= 1.
The medians AD, BE, CF are therefore concurrent. Their intersection isthe centroid G of the triangle.
CB
A
GF
D
E
Consider the line BGE intersecting the sides of triangle ADC. By theMenelaus theorem,
−1 =AG
GD· DB
BC· CE
EA=
AG
GD· −1
2· 11.
It follows that AG : GD = 2 : 1.The centroid of a triangle divides each median in the ratio 2:1.
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The incenter
Let X , Y , Z be points on BC, CA, AB such that AX , BY , CZ bisectangles BAC, CBA and ACB respectively. Then
AZ
ZB=
b
a,
BX
XC=
c
b,
CY
Y A=
a
c.
CB
A
IZ
X
Y
It follows that
AZ
ZB· BX
XC· CY
Y A=
b
a· cb· ac= +1,
and AX , BY , CZ are concurrent. Their intersection is the incenter of thetriangle.
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Example
Given a point P , let the lines AP , BP , CP intersectBC, CA, AB respectively at X , Y , Z.Construct the circle through X , Y , Z,to intersect the lines BC, CA, AB again at X ′, Y ′, Z ′.Then the lines AX ′, BY ′, CZ ′ are concurrent.
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Example
Suppose two cevians, each through a vertex of a triangle, trisect each other.Show that these are medians of the triangle.
A
B C
YZ
P
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Solution
Suppose two cevians, each through a vertex of a triangle, trisect each other.Show that these are medians of the triangle.
A
B C
YZP
Given: Triangle ABC with cevians BY and CZ intersecting at Psuch that P trisects BY and CZ.To prove: BY and CZ are medians,i.e., Y and Z are midpoints of CA and AB respectively.
Proof. (1) Since P is a trisection of BY and CQ,ZCCP = −3
p for p = 1 or 2,PYY B = −q
3 for q = 1 or 2.(2) Applying Menelaus’ theorem to triangle BPZ with transversal CY A,we have. . .
26
(3) Similarly, by applying Menelaus’ theorem to triangleCPY with transver-sal AZB, we conclude that Y is the midpoint of CA, and BY is also amedian.
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Example
Let AX , BY , CZ be cevians of �ABC intersecting at a point P .(i) Show that if AX bisects angle A, and BX · CY = XC · BZ, then
�ABC is isosceles.(ii) Show if if AX , BY , CZ are bisectors, and BP · ZP = BZ · AP ,
then �ABC is a right triangle.
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Example
ABC is an isosceles triangle with β = γ = 40◦. Y and Z are points on CA
and AB respectively such that BY bisects angle B and ∠ZCB = 30◦. LetP be the intersection of BY and CZ. Show that BP = AB.
A
B C
Y
Z
P
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Example
Let ABC be a triangle with A = 40◦, B = 60◦. Let E and F be pointslying on the sides AC and AB respectively, such that ∠EBC = 40◦ and∠FCB = 70◦. Let BE and CF intersect at P . Show that AP is perpendic-ular to BC.
A
B C
EP
F
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Example
Let A,B,C,D,E, F be six consecutive points on a circle. Show that thechords AD, BE, CF are concurrent if and only if AB · CD · EF = BC ·DE · FA.
A
B
C
D
E
F
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Example
A1A2A3 · · ·A11A12 is a regular 12-gon. Show that the diagonals A1A5,A3A6, and A4A8 are concurrent.
A1
A2
A3
A4
A5
A6
A7
A8
A9
A10
A11
A12
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Kiepert perspectors
If similar isosceles triangles A′BC, B′CA and C ′AB (of base angle θ)are constructed on the sides of triangle ABC,either all externally or all internally,the lines AA′, BB′, and CC ′ are concurrent.
A
B C
C′
A′
B′
P
α1 α2
θ
β1
β2
γ2
γ1
θ
θ
θ θ
θ
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Proof
Proof. Applying the law of sines to triangles ABA′ and ACA′, we have
sinα1
sinα2=
sinα1
sin(β + θ)· sin(β + θ)
sin(γ + θ)· sin(γ + θ)
sinα2
=BA′
AA′ ·sin(β + θ)
sin(γ + θ)· AA
′
CA′
=sin(β + θ)
sin(γ + θ).
Likewise, sinβ1
sinβ2= sin(γ+θ)
sin(α+θ) and sin γ1sin γ2
= sin(α+θ)sin(β+θ) . From these
sinα1
sinα2· sin β1sin β2
· sin γ1sin γ2
= +1.
and the lines AA′, BB′, CC ′ are concurrent.
The point of intersection is called the Kiepert perspector K(θ). In par-ticular, it is called(1) a Fermat point if θ = ±60◦,(2) a Napoleon point if θ = ±30◦,(3) a Vecten point if θ = ±45◦.
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Trigonmetric version of the Ceva Theorem
Let X be a point on the side BC of triangle ABC such that the directedangles ∠BAX = α1 and ∠XAC = α2. By the sine formula,
BX
XC=
BX/AX
XC/AX=
sinα1/ sin β
sinα2/ sin γ=
sin γ
sin β· sinα1
sinα2=
c
b· sinα1
sinα2.
X
A
B C
Z
Y
α1 α2
β1
β2
γ2
γ1
Likewise, if Y and Z be points on the lines CA, AB respectively, with∠CBY = β1, ∠Y BA = β2 and ∠ACZ = γ1, ∠ZCB = γ2, we have
CY
Y A=
a
c· sin β1sin β2
,AZ
ZB=
b
a· sin γ1sin γ2
.
These lead to the following trigonometric version of the Ceva theorem.Yiu: Advanced Euclidean Geometry 2016
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Theorem 0.3 (Ceva). The lines AX , BY , CZ are concurrent if and only if
sinα1
sinα2· sin β1sin β2
· sin γ1sin γ2
= +1.
Proof.AZ
ZB· BX
XC· CY
Y B=
sinα1
sinα2· sin β1sin β2
· sin γ1sin γ2
.
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