*CevaÕs
Theorem*

* *

*By Courtney
Cody*

In this assignment, we will
investigate, prove, and extend CevaÕs Theorem.

LetÕs begin by constructing a
diagram to represent CevaÕs Theorem.
Consider any triangle ABC.
Select a point P inside the triangle and draw lines *AP, BP*, and *CP*
extended to their intersections with the opposite sides in points *D, E*, and *F*
respectively. We wish to explore
the relationship between the products and , or the product of the left-hand-sides (green) and the
right-hand sides (orange) of each side of triangle *ABC*.

To begin our investigation,
we will the built-in measure and calculation tools to explore the relationship
between the two products.

As we can see in the diagram
above, the products and appear to be
equal. Click HERE
to move the points of the triangle around to see if the products remain the
same.

**THEOREM**

CevaÕs theorem states that if points *D, E*, and *F* are on the sides *AB*, *BC* and *AC* of a triangle *ABC* then the lines *AE, BF *and *CD* are concurrent if and only if the
product of the ratios

.

**PROOF**

** **

We will begin by assuming the
following diagram for some point *P*
inside the triangle:

Suppose we create a line
parallel to *BC* through *A* and extend the segments *CF* and *BE*
beyond the triangle to form points *G*
and *H* respectively as the lines
intersect the parallel line through *A*. Several sets of similar triangles form,
four of which we will pay particular attention to, which are described below.

(1) *Triangle AFG is
similar to triangle CBF.*

The two green triangles
pictured in the diagram above are similar because opposite angles *AFG* and *BFC* are
congruent, *AGF* and *FCB* are
alternate interior angles, and thus *GAF* and *FBC* are
congruent angles because all other corresponding angles are congruent and by
the triangle sum theorem the third corresponding angles in each of the two
triangles must be congruent.
Therefore, triangle *AFG* is
similar to triangle *CBF*.

Since *AFG~BFC*, we can set up the following proportion:

(2) *Triangle BDP is
similar to triangle HAP.*

By a similar method as in
(1), triangle *BDP* is similar to
triangle *HAP*. Since *BDP~HAP*, we can set up the following proportion:

(3) *Triangle AHE is
similar to triangle CBE.*

By a similar method as in
(1), triangle *AHE* is similar to
triangle *CBE*. Since *AHE~CBE*, we can set up the following proportion:

(4) *Triangle APG is
similar to triangle DPC.*

By a similar method as in
(1), triangle *APG* is similar to
triangle *DPC*. Since *APG~DPC*, we can set up the following proportion:

Hence, after exploring the
different similar triangles, we have the following four proportions:

(a) (b)
(c) (d)

Since the ratio is common to
both (b) and (d), then we know that .

Since *G *is not a part of the original triangle, we want to
eliminate *AG* from the above
proportion. Hence, we can use (a)
to solve for *AG* and substitute.

Since , then .

Similarly, since *H* is not a part of the original triangle, we want to
eliminate *AH* from the above
proportion. Hence, we can use (c)
to solve for *AH* and substitute.

Since , then .

Now, by the reflexive
property *CB=BC, DC=CD, DB=BD, *and*
BF=FB.*

* *

Hence, making these
substitutions and rearranging the terms, we have

Thus, .

** **

**PROOF OF THE CONVERSE OF
CEVAÕS THEOREM**

Consider any triangle *ABC* and let *D, E*, and *F* lie on segments *BC,
AC*, and *AB* respectively, all of which satisfy

.

Let P be the intersection
inside triangle *ABC* of the
segments *CF* and *BE*. Also,
let *DÕ* be the intersection of *AP* and *BC*.

Since *AD, BE* and *CF* are concurrent, then we know from our assumption that

However, we assumed

.

Equating these two equations,
we obtain:

Thus, *D=DÕ*, which implies that *AD, BE* and *CF* are concurrent. Therefore, CevaÕs theorem holds.

** **

** **

**CONCURRENCY OF THREE
MEDIANS**

Suppose *P *is the centroid of triangle *ABC*.

How could we use CevaÕs
theorem to prove concurrency of the medians? If we recall the definition of a centroid of a triangle, it is
the point of intersection of the triangleÕs three medians. In relating this definition to CevaÕs
theorem, this means that points *D, E*,
and *F* are the midpoints of
segments *BC, AC*, and *AB* respectively.
Hence, *AF=FB, BD=CD*, and *CE=AE*. Now,
CevaÕs theorem states that segments *AD, BE* and *CF*
are concurrent if and only if the product of the ratios

.

In substituting the values of
the aforementioned equivalent segments, we have

Therefore, since the ratio of
the products is one, then the three medians, namely *AD, BE* and *CF*, are concurrent.

** **

** **

**GENERALIZATION**

** **

Suppose we want to explore a
generalization of CevaÕs theorem such that *P* can lie outside of triangle *ABC*. In
order for *P* to lie outside the
triangle, we must construct our diagram using lines rather than segments. Click HERE
to see a working GSP sketch of this case.

** **

** **

Therefore, we see from the
diagram and the GSP sketch that even when *P* is not contained within triangle *ABC*, CevaÕs theorem still holds since the ratio of the
products is still one.