6. Theorem of Ceva, Menelaus and Van Aubel.

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Oct 8, 2013 (3 years and 10 months ago)

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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