6.Theorem of Ceva,Menelaus and Van Aubel.

Theorem 1 (Menelaus).If A

1

;B

1

;C

1

are points on the sides BC;CA

and AB of a triangle ABC;then the points are collinear if and only if

jA

1

Bj

jA

1

Cj

:

jB

1

Cj

jB

1

Aj

:

jC

1

Aj

jC

1

Bj

= 1:

Figure 1:

Proof Assume points are collinear.

First drop perpendiculars AA

0

;BB

0

and CC

0

from the vertices A;B;C to the line A

1

B

1

C

1

:

Then since AA

0

;BB

0

and CC

0

are perpendic-

ular to A

1

B

1

;they are parallel (Figure 1).

Thus we get the following equalities of ratios

jA

1

Bj

jA

1

Cj

=

jBB

0

j

jCC

0

j

;

jB

1

Cj

jB

1

Aj

=

jCC

0

j

jAA

0

j

and

jC

1

Aj

jC

1

Bj

=

jAA

0

j

jBB

0

j

:

Multiplying these we get the required re-

sult.

Conversely,suppose

jA

1

Bj

jA

1

Cj

:

jB

1

Cj

jB

1

Aj

:

jC

1

Aj

jC

1

Bj

= 1:

Now suppose lines BC and B

1

C

1

meet at the point A

00

:Then

jA

00

Bj

jA

00

Cj

:

jB

1

Cj

jB

1

Aj

:

jC

1

Aj

jC

1

Bj

= 1:

Thus

jA

1

Bj

jA

1

Cj

=

jA

00

Bj

jA

00

Cj

;

1

and so we conclude that the point A

00

on the line BC coincides with the point

A

1

:Thus the points A

1

;B

1

and C

1

are collinear.

De¯nition 1 A line segment joining a vertex of a triangle to any

given point on the opposite side is called a Cevian.

Theorem 2 (Ceva) Three Cevians AA

1

;BB

1

and CC

1

of a triangle ABC

(Figure 2) are concurrent if and only if

jBA

1

j

jA

1

Cj

:

jCB

1

j

jB

1

Aj

:

jAC

1

j

jC

1

Bj

= 1.

Figure 2:

Proof First assume that the Cevians are con-

current at the point M:

Consider the triangle AA

1

C and apply Menelaus'the-

orem.Since the points B

1

;M and B are collinear,

jB

1

Cj

jB

1

Aj

:

jMAj

jMA

1

j

:

jBA

1

j

jBCj

= 1:::(a)

Now consider the triangle AA

1

B:The points C

1

;M;C

are collinear so

jC

1

Aj

jC

1

Bj

:

jCBj

jCA

1

j

:

jMA

1

j

jMAj

= 1:::(b)

Multiply both sides of equations (a) and (b) to get re-

quired result.

Conversely,suppose the two Cevians AA

1

and BB

1

meet at P and that

the Cevian from the vertex C through P meets side AB at C

0

:Then we have

jBA

1

j

jA

1

Cj

:

jCB

1

j

jB

1

Aj

:

jAC

0

j

jC

0

Bj

= 1:

By hypothesis,

jBA

1

j

jA

1

Cj

:

jCB

1

j

jB

1

Aj

:

jAC

1

j

jC

1

Bj

= 1:

Thus

jAC

1

j

jC

1

Bj

=

jAC

0

j

jC

0

Bj

;

and so the two points C

1

and C

0

on the line segment AB must coincide.The

required result follows.

2

Theorem 3 (van Aubel) If A

1

;B

1

;C

1

are interior points of the sides BC;CA

and AB of a triangle ABC and the corresponding Cevians AA

1

;BB

1

and

CC

1

are concurrent at a point M (Figure 3),then

jMAj

jMA

1

j

=

jC

1

Aj

jC

1

Bj

+

jB

1

Aj

jB

1

Cj

:

Figure 3:

Proof Again,as in the proof of Ceva's theo-

rem,we apply Menelaus'theorem to the triangles AA

1

C

and AA

1

B:

In the case of AA

1

C;we have

jB

1

Cj

jB

1

Aj

:

jMAj

jMA

1

j

:

jBA

1

j

jBCj

= 1;

and so

jB

1

Aj

jB

1

Cj

=

jMAj

jMA

1

j

:

jBA

1

j

jBCj

:::(c)

For the triangle AA

1

B;we have

jC

1

Aj

jC

1

Bj

:

jCBj

jCA

1

:

jMA

1

j

jMAj

= 1;

and so

jC

1

Aj

jC

1

Bj

=

jMAj

jMA

1

j

:

jCA

1

j

jBCj

:::(d)

Adding (c) and (d) we get

jB

1

Aj

jB

1

Cj

+

jC

1

Aj

jC

1

Bj

=

jMAj

jMA

1

jjBCj

fjBA

1

j +jA

1

Cjg =

jMAj

jMA

1

j

;

as required.

Examples

1.Medians AA

1

;BB

1

and CC

1

intersect at the centroid G and then

jGAj

jGA

1

j

= 2;

since

1 =

jA

1

Bj

jA

1

Cj

=

jB

1

Cj

jB

1

Aj

=

jC

1

Aj

jC

1

Bj

:

3

2.The angle bisectors in a triangle are concurrent at the incentre I of

the triangle.Furthermore,if A

3

;B

3

and C

3

are the points on the sides

BC;CA and AB where the bisectors intersect these sides (Figure 4),

then

jA

3

Bj

jA

3

Cj

=

c

b

;

jB

3

Cj

jB

3

Aj

=

a

c

and

jC

3

Aj

jC

3

Bj

=

b

a

:

Then

jIAj

jIA

3

j

=

jC

3

Aj

jC

3

Bj

+

jB

3

Aj

jB

3

Cj

=

b

a

+

c

a

=

b +c

a

:

Figure 4:

3.Let AA

2

;BB

2

and CC

2

be the altitudes of a triangle ABC:They are

concurrent at H;the orthocentre of ABC (Figure 5.)

We have

jA

2

Bj

jA

2

Cj

=

jAA

2

j cot(

b

B)

jAA

2

j cot(

b

C)

=

tan(

b

C)

tan(

b

B)

and similarly

jB

2

Cj

jB

2

Aj

=

tan(

b

A)

tan(

b

C)

;

jC

2

Aj

jC

2

Bj

=

tan(

b

B)

tan(

b

C)

:

4

Figure 5:

Multiplying the 3 ratios,we get concurrency of the altitudes.Further-

more,

jHAj

jHA

2

j

=

jC

2

Aj

jC

2

Bj

+

jB

2

Aj

jB

2

Cj

=

tan(

b

B)

tan(

b

A)

+

tan(

b

C)

tan(

b

A)

=

tan(

b

B) +tan(

b

C)

tan(

b

A)

=

sin(

b

B +

b

C):cos(

b

A)

cos(

b

B) cos(

b

C) sin(

b

A)

=

sin(180

±

¡

b

A) cos(

b

A)

cos(

b

B) cos(

b

C) sin(

b

A)

=

cos(

b

A)

cos(

b

B) cos(

b

C)

:

Lemma 1 Let ABC be a triangle and A

1

a point on the side BC so

that

jA

1

Bj

jA

1

Cj

=

°

¯

Let X and Y be points on the sides AB and AC respectively and let M be

the point of intersection of the line segments XY and AA

1

(Figure 6).Then

¯(

jXBj

jXAj

) +°(

jY Cj

jY Aj

) = (¯ +°)(

jA

1

Mj

jMAj

):

Figure 7:

Proof First suppose that

XY is parallel to the side BC:Then

jXBj

jXAj

=

jY Cj

jY Aj

=

jMA

1

j

jMAj

;

5

Figure 6:

and so result is true for any ¯ and

°:

Now suppose the lines XY and BC in-

tersect at a point Z:

Consider the triangle AA

1

B (Figure

7).Since M;X and Z are collinear,

jY Cj

jY Aj

:

jMAj

jMA

1

j

:

jZA

1

j

jZCj

= 1:

Then ¯(

jXBj

jXAj

) +°(

jY Cj

jY Aj

)

= ¯(

jMA

1

jjZBj

jMAjjZA

1

j

) +°(

jMA

1

jjZCj

jMAjjZA

1

j

)

=

jMA

1

j

jMAjjZA

1

j

f¯jZBj +°jZCjg

=

jMA

1

j

jMAjjZA

1

j

f¯jZA

1

j ¡¯jBA

1

j +°jZA

1

j +°jA

1

Cjg

= (¯ +°)

jMA

1

j

jMAjjZA

1

j

:jZA

1

j;

since

jBA

1

j

j

A

1

C

j

=

°

¯

;

= (¯ +°)

jMA

1

j

jMAj

;as required:

6

Theorem 4 Let ABC be a triangle with three cevians AA

1

;BB

1

and

CC

1

intersecting at a point M (Figure 8).

Figure 8:

Furthermore suppose

jA

1

Bj

jA

1

Cj

=

°

¯

;

jB

1

Cj

jB

1

Aj

=

®

°

and

jC

1

Aj

jC

1

Bj

=

¯

®

:

If X and Y are points on the sides AB and AC then the point M belongs to

the line segment XY if and only if

¯(

jXBj

jXAj

) +°(

jY Cj

jY Aj

) = ®:

Proof By van Aubel's theorem:

jAMj

jA

1

Mj

=

jC

1

Aj

jC

1

Bj

+

jB

1

Aj

jB

1

Cj

=

¯

®

+

°

®

=

¯ +°

®

:

Nowsuppose M belongs to the line segment XY:Then by the previous lemma

¯(

jXBj

jXAj

) +°(

jY Cj

jY Aj

) = (¯ +°)

jA

1

Mj

jMAj

= (¯ +°)(

®

¯ +°

) = ®;as required:

For converse,suppose XY and AA

1

intersect in point M

0

:We will show that

M

0

coincides M:

By the lemma,

7

¯(

jXBj

jXAj

) +°(

jY Cj

jY Aj

) = (¯ +°)(

jA

1

M

0

j

jM

0

Aj

):

By hypothesis,we have

¯(

jXBj

jXAj

) +°(

jY Cj

jY Aj

) = ®:

Thus

jA

1

Mj

jAMj

=

®

¯ +°

;

and so M and M

0

coincide.Thus M must lie on the line segment XY:

Corollary 1 If G is the centroid of the triangle ABC and so ® = ¯ =

° = 1;then G belongs to the line segment XY if and only if

jXBj

jXAj

+

jY Cj

jY Aj

= 1:

Corollary 2 If I is the incentre of the triangle ABC then the values

of ®;¯ and ° are given in terms of the sidelengths of the triangle as

® = a;¯ = b and ° = c:

Thus I belongs to XY if and only if

b(

jXBj

jXAj

) +c(

jY Cj

jY Aj

) = a:

Corollary 3 If H is the orthocentre of the triangle ABC then the

ratios on the sides are given by

® = tan(

b

A);¯ = tan(

b

B) and ° = tan(

b

C:)

Then we get that H belongs to the line segment XY if and only if

(tan(

b

B))(

jXBj

jXAj

) +(tan(

b

C))(

jY Cj

jY Aj

) = tan(

b

A):

We also get the following result which was a question on the 2006 Irish

Invervarsity Mathematics Competition.

Theorem 5 Let ABC is a triangle and let X and Y be points on the

sides AB and AC respectively such that the line segment XY bisects the area

of ABC and the points X and Y bisects the perimeter (Figure 9).Then the

incentre I belongs to the line segment XY.

8

Figure 9:

Proof Let x = jAXj and y = jAY j:

Then

x +y =

a +b +c

2

:::(a)

where a;b and c are lengths of sides.

Furthermore,

1

2

=

area(AXY )

area(ABC)

=

xy sin(

b

A)

bc sin(

b

A)

;

so

xy =

bc

2

:::(b):

Consider b(

jXBj

jXAj

) +c(

jY Cj

jY Aj

)

= b(

c ¡x

x

) +c(

b ¡y

y

)

= b(

1

x

+

1

y

) ¡b ¡c

= bc(

a +b +c

2

:

2

bc

) ¡b ¡c

= a:

Thus by Corollary 2,incentre I belongs to the line XY:

Theorem 6 Let ABC be an equilateral triangle and X;Y and Z points

on the sides BC;CA and AB respectively (Figure 10).Then the minimum

value of

jZXj

2

+jXY j

2

+jY Zj

2

is attained when X;Y;Z are the midpoints of the sides.

9

Figure 10:

Proof Consider

1

3

fjZXj

2

+jXY j

2

+jY Zj

2

g

We have

1

3

fjZXj

2

+jXY j

2

+jY Zj

2

g

¸ (

jZXj +jXY j +jY Zj

2

)

2

;

by Cauchy ¡Schwarz inequality;

¸ (

jA

1

B

1

j +jB

1

C

1

j +jC

1

A

1

j

3

)

2

;

where A

1

B

1

C

1

is the orthic triangle of ABC.(This result was proved in

chapter 5 on orthic triangles.)

If l is the common value of the sides of ABC then the orthic triangle A

1

B

1

C

1

is also equilateral and sidelengths are

l

2

:Thus

(

jA

1

B

1

j +jB

1

C

1

j +jC

1

A

1

j

3

)

2

= jA

1

B

1

j

2

=

jA

1

B

1

j

2

+jB

1

C

1

j

2

+jC

1

A

1

j

2

3

:

The required result follows.

10

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