CEVAŐS THEOREM
BY
SHADRECK S
CHITSONGA
PROOFS
Here we will look at the proofs to the conjectures we
made earlier on. We will look at three situations.
1. When P is the centroid of triangle ABC
Given: with P as the
centroid of the triangle, such
that and , produced meet the sides of
at E, F and D
respectively.
To prove: that
Proof: P is the centroid
(given)
D,E, and F are
the mid-points of sides BC,CA, and
AB respectively.
=, = , and = . (i)
Now let us consider the products (AF)(BD)(CE) and
(BF)(CD)(AE).
Using (i) we see that the products are equal to each
other.
, holds.
2. When P is the orthocenter of the triangle
Given: with P as the
orthocenter of the triangle ,
such that and , produced meet the
sides
of at E,F and
D respectively.
To prove : that
Proof: P is
the orthocenter (given)
A line from the vertex of a triangle that passes
through the
orthocenter is perpendicular to the
opposite side.
The angles AFP, BFP, BDP, PDC, CEP, PEA are all
equal to 90ˇ , therefore equal to each other.
In triangles BPF and
CPE,
= 90ˇ (proved).
(vertically
opposite angles).
(angle sum of triangle =180ˇ)
is similar to .
(i)
Similarly we can show that is similar
to
.
(ii)
Also is similar to .
(iii),
From (i) and from
(iii)
(iv)
From (iii) , and we can substitute
this in (iv).
Dividing both sides by AP and some
rearrangements we end up
with: (v)
You will notice that (v) is equivalent to
3. When P is the incenter
Given: with P as the
incenter triangle ,
such that and , produced meet the
sides of at E,F and
D respectively.
To prove: that
Proof : Before doing this proof
we need some background
information. We need to make use of another
theorem. This theorem states that:
The bisector of an
interior angle of a triangle divides the opposite side into segments
proportional to the adjacent sides. Look at the GSP in figure 1.
Figure
1.
NOTE: This is not a proof for this theorem. We are
just using the result here.
Using the results in figure 1,
we now make use of the ratios
(BD/DC)=(AB/AC),
(AC/BC)=(AF/FB), and (AE/EC)=(AB/BC).
We can rearrange these ratios
as
(i) AB={(BD X AC)/DC)} (ii) AB = {(AE X BC)/ CE)}
(iii) AC = {(FA X BC)/BF}
Using (i) and (ii) we have {(BD
X AC)/DC)}= {(AE X BC)/ CE)} (iv).
Now substituting (iii) in (iv)
we have:
{(BD )(FA X BC)/(BF X DC)} =
{(AE X BC)/(CE)}
this simplifies to (BD) (FA)
(CE) = (AE) (BF) (DC).
Therefore {(BD) (FA) (CE)}/
{(AE) (BF) (DC)} = 1.
4. When P is an arbitrary point inside the triangle.
Given: with P as an
arbitrary point inside the triangle ,
such that and , produced meet the sides of
at E,F and
D respectively.
To prove: that
Construction: Draw parallel to , to meet in Y.
Draw parallel to to meet in X.
Proof:
is
parallel to (construction) .
also
We
can write also
(i)
Equation (i) can also be expressed as
is parallel to (construction)
also
We can write also
(ii)
Equation(ii) can
also be expressed as
Because equations (i) and (ii) are both equal to 1,
We can write
After simplifying we end up with:
, which is
equivalent to
*NOTE: I considered including the converse after reading your
comments.
To prove the
converse, we must show that if ,
holds then there must be a a common point P where all the three Cevians pass
through.
GIVEN:
, where D, E, and F are points on the sides
BC, AC, and AB
respectively.
TO
PROVE: That segments BE,
AD and CF meet a common point( say P)
PROOF:
, , and are the sides of
triangle ABC, therefore they
are all not parallel to each
other.
We will assume that the segments BE, AD, and CF are not
parallel to each other.
With this assumption, then two of the segments must intersect
at a point.
Let
us consider segments BE and CF and let them intersect
in P. This means that the two segments contain the point P.
Now we can draw a line AP. If we assume that that line AP is
not parallel to segment BC, then the two must intersect at
point. Let us call this point DŐ on the segment BC.
If this is the case then we have segments BE, CF and ADŐ
passing
through a common point P.
This then means that , must hold.
But we have been given that .
We can see that , which means that DŐ and D are
the same point.
We can therefore conclude that all the three segments AD,
BE, and CF contain the same point P, therefore they are
concurrent.