6. Theorem of Ceva, Menelaus and Van Aubel.
Theorem 1
(Menelaus). If A1 , B1 , C1 are points on the sides BC, CA
and AB of a triangle ABC, then the points are collinear if and only if
|A1 B| |B1 C| |C1 A|
.
.
= 1.
|A1 C| |B1 A| |C1 B|
Proof
Assume points are collinear.
First drop perpendiculars AA0 , BB 0 and CC 0
from the vertices A, B, C to the line A1 B1 C1 .
Then since AA0 , BB 0 and CC 0 are perpendicular to A1 B1 , they are parallel (Figure 1).
Thus we get the following equalities of ratios
|A1 B|
|BB 0 |
=
,
|A1 C|
|CC 0 |
and
|B1 C|
|CC 0 |
=
|B1 A|
|AA0 |
|C1 A|
|AA0 |
=
.
|C1 B|
|BB 0 |
Multiplying these we get the required result.
Conversely, suppose
|A1 B| |B1 C| |C1 A|
.
.
= 1.
|A1 C| |B1 A| |C1 B|
Now suppose lines BC and B1 C1 meet at the point A00 . Then
|A00 B| |B1 C| |C1 A|
.
.
= 1.
|A00 C| |B1 A| |C1 B|
|A1 B|
|A00 B|
Thus
= 00 ,
|A1 C|
|A C|
1
Figure 1:
and so we conclude that the point A00 on the line BC coincides with the point
A1 . Thus the points A1 , B1 and C1 are collinear.
Definition 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 AA1 , BB1 and CC1 of a triangle ABC
(Figure 2) are concurrent if and only if
|BA1 | |CB1 | |AC1 |
.
.
= 1.
|A1 C| |B1 A| |C1 B|
Proof
First assume that the Cevians are concurrent at the point M.
Consider the triangle AA1 C and apply Menelaus’ theorem. Since the points B1 , M and B are collinear,
|B1 C| |M A| |BA1 |
.
.
=1
|B1 A| |M A1 | |BC|
. . . (a)
Now consider the triangle AA1 B. The points C1 , M, C
are collinear so
|C1 A| |CB| |M A1 |
.
.
=1
. . . (b)
|C1 B| |CA1 | |M A|
Figure 2:
Multiply both sides of equations (a) and (b) to get required result.
Conversely, suppose the two Cevians AA1 and BB1 meet at P and that
the Cevian from the vertex C through P meets side AB at C 0 . Then we have
|BA1 | |CB1 | |AC 0 |
.
.
= 1.
|A1 C| |B1 A| |C 0 B|
By hypothesis,
|BA1 | |CB1 | |AC1 |
.
.
= 1.
|A1 C| |B1 A| |C1 B|
|AC1 |
|AC 0 |
Thus
= 0 ,
|C1 B|
|C B|
and so the two points C1 and C 0 on the line segment AB must coincide. The
required result follows.
2
Theorem 3 (van Aubel) If A1 , B1 , C1 are interior points of the sides BC, CA
and AB of a triangle ABC and the corresponding Cevians AA1 , BB1 and
CC1 are concurrent at a point M (Figure 3), then
|M A|
|C1 A| |B1 A|
=
+
.
|M A1 |
|C1 B| |B1 C|
Proof
Again, as in the proof of Ceva’s theorem, we apply Menelaus’ theorem to the triangles AA1 C
and AA1 B.
In the case of AA1 C, we have
|B1 C| |M A| |BA1 |
.
.
= 1,
|B1 A| |M A1 | |BC|
and so
|B1 A|
|M A| |BA1 |
=
.
|B1 C|
|M A1 | |BC|
. . . (c)
Figure 3:
For the triangle AA1 B, we have
|C1 A| |CB| |M A1 |
.
.
= 1,
|C1 B| |CA1 |M A|
and so
|C1 A|
|M A| |CA1 |
=
.
|C1 B|
|M A1 | |BC|
. . . (d)
Adding (c) and (d) we get
|B1 A| |C1 A|
|M A|
+
=
{|BA1 | + |A1 C|}
|B1 C| |C1 B|
|M A1 ||BC|
=
|M A|
,
|M A1 |
as required.
Examples
1. Medians AA1 , BB1 and CC1 intersect at the centroid G and then
|GA|
= 2,
|GA1 |
since
1=
|A1 B|
|B1 C|
|C1 A|
=
=
.
|A1 C|
|B1 A|
|C1 B|
3
2. The angle bisectors in a triangle are concurrent at the incentre I of
the triangle. Furthermore, if A3 , B3 and C3 are the points on the sides
BC, CA and AB where the bisectors intersect these sides (Figure 4),
then
|A3 B|
c |B3 C|
a
|C3 A|
b
= ,
= and
= .
|A3 C|
b |B3 A|
c
|C3 B|
a
|C3 A| |B3 A|
|IA|
=
+
Then
|IA3 |
|C3 B| |B3 C|
b
c
b+c
= + =
.
a a
a
Figure 4:
3. Let AA2 , BB2 and CC2 be the altitudes of a triangle ABC. They are
concurrent at H, the orthocentre of ABC (Figure 5.)
We have
b
|A2 B|
|AA2 | cot(B)
=
b
|A2 C|
|AA2 | cot(C)
b
tan(C)
=
b
tan(B)
and similarly
b
tan(A)
|B2 C|
=
,
b
|B2 A|
tan(C)
b
tan(B)
|C2 A|
=
.
b
|C2 B|
tan(C)
4
Figure 5:
Multiplying the 3 ratios, we get concurrency of the altitudes. Furthermore,
|HA|
|HA2 |
=
=
=
Lemma 1
that
|C2 A| |B2 A|
+
|C2 B| |B2 C|
b + tan(C)
b
tan(B)
=
=
b
b
tan(B)
tan(C)
+
b
b
tan(A)
tan(A)
=
b
cos(A)
.
b cos(C)
b
cos(B)
b
tan(A)
b + C).
b cos(A)
b
sin(B
b cos(C)
b sin(A)
b
cos(B)
b cos(A)
b
sin(180◦ − A)
b cos(C)
b sin(A)
b
cos(B)
Let ABC be a triangle and A1 a point on the side BC so
|A1 B|
γ
=
|A1 C|
β
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 AA1 (Figure 6). Then
β(
|XB|
|Y C|
|A1 M |
) + γ(
) = (β + γ)(
).
|XA|
|Y A|
|M A|
Proof
First suppose that
XY is parallel to the side BC. Then
|XB|
|Y C|
|M A1 |
=
=
,
|XA|
|Y A|
|M A|
5
Figure 7:
Figure 6:
and so result is true for any β and
γ.
Now suppose the lines XY and BC intersect at a point Z.
Consider the triangle AA1 B (Figure
7). Since M, X and Z are collinear,
|Y C| |M A| |ZA1 |
.
.
= 1.
|Y A| |M A1 | |ZC|
Then
=
=
|XB|
|Y C|
) + γ(
)
|XA|
|Y A|
|M A1 ||ZB|
|M A1 ||ZC|
β(
) + γ(
)
|M A||ZA1 |
|M A||ZA1 |
|M A1 |
{β|ZB| + γ|ZC|}
|M A||ZA1 |
β(
|M A1 |
{β|ZA1 | − β|BA1 | + γ|ZA1 | + γ|A1 C|}
|M A||ZA1 |
|M A1 |
.|ZA1 |,
=
(β + γ)
|M A||ZA1 |
|BA1 |
γ
since
= ,
|A1 C|
β
|M A1 |
= (β + γ)
,
as required.
|M A|
=
6
Theorem 4
Let ABC be a triangle with three cevians AA1 , BB1 and
CC1 intersecting at a point M (Figure 8).
Figure 8:
Furthermore suppose
|A1 B|
γ |B1 C|
α
|C1 A|
β
= ,
=
and
= .
|A1 C|
β |B1 A|
γ
|C1 B|
α
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
β(
Proof
|XB|
|Y C|
) + γ(
) = α.
|XA|
|Y A|
By van Aubel’s theorem:
|AM |
|A1 M |
|C1 A| |B1 A|
+
|C1 B| |B1 C|
β
γ
β+γ
= + =
.
α α
α
=
Now suppose M belongs to the line segment XY. Then by the previous lemma
β(
|XB|
|Y C|
) + γ(
)
|XA|
|Y A|
|A1 M |
|M A|
α
) = α,
= (β + γ)(
β+γ
= (β + γ)
as required.
For converse, suppose XY and AA1 intersect in point M 0 . We will show that
M 0 coincides M.
By the lemma,
7
β(
|XB|
|Y C|
|A1 M 0 |
) + γ(
) = (β + γ)(
).
|XA|
|Y A|
|M 0 A|
By hypothesis, we have
β(
|XB|
|Y C|
) + γ(
) = α.
|XA|
|Y A|
Thus
|A1 M |
α
=
,
|AM |
β+γ
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 α = β =
γ = 1, then G belongs to the line segment XY if and only if
|XB| |Y C|
+
= 1.
|XA| |Y A|
Corollary 2
If I is the incentre of the triangle ABC then the values
of α, β and γ are given in terms of the sidelengths of the triangle as
α = a, β = b and γ = c.
Thus I belongs to XY if and only if
b(
|XB|
|Y C|
) + c(
) = a.
|XA|
|Y A|
Corollary 3
If H is the orthocentre of the triangle ABC then the
ratios on the sides are given by
b
b and γ = tan(C.)
b
α = tan(A),
β = tan(B)
Then we get that H belongs to the line segment XY if and only if
b
(tan(B))(
|XB|
b |Y C| ) = tan(A).
b
) + (tan(C))(
|XA|
|Y 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
Proof
Let x = |AX| and y = |AY |.
Then
x+y =
a+b+c
2
. . . (a)
where a, b and c are lengths of sides.
Furthermore,
b
1
area(AXY )
xy sin(A)
=
=
,
b
2
area(ABC)
bc sin(A)
so
xy =
Consider
bc
2
Figure 9:
. . . (b).
|XB|
|Y C|
) + c(
)
|XA|
|Y A|
b−y
c−x
) + c(
)
= b(
x
y
1 1
= b( + ) − b − c
x y
a+b+c 2
= bc(
. )−b−c
2
bc
= a.
b(
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
|ZX|2 + |XY |2 + |Y Z|2
is attained when X, Y, Z are the midpoints of the sides.
9
Figure 10:
Proof
Consider
We have
1
{|ZX|2 + |XY |2 + |Y Z|2 }
3
1
{|ZX|2 + |XY |2 + |Y Z|2 }
3
|ZX| + |XY | + |Y Z| 2
≥(
),
2
by Cauchy − Schwarz inequality,
|A1 B1 | + |B1 C1 | + |C1 A1 | 2
),
≥(
3
where A1 B1 C1 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 A1 B1 C1
l
is also equilateral and sidelengths are . Thus
2
(
|A1 B1 | + |B1 C1 | + |C1 A1 | 2
) = |A1 B1 |2
3
|A1 B1 |2 + |B1 C1 |2 + |C1 A1 |2
=
.
3
The required result follows.
10