** **

**CevaŐs Theorem **

**By Mary
Negley**

** **

For
this write-up, I will explore CevaŐs Theorem, which states that given a
triangle,
the three lines,,
and
intersect at point *iff
*. A GSP
sketch is given below.

An
interesting fact about point *P* is
that it may be anywhere on the plane. I will look at *P* in the following cases: (1) when *P* is the centroid, (2) when *P* is the orthocenter, (3) when *P* is the incenter, and (4) when *P* is the circumcenter.

Case1:
*P* is the centroid. By definition of centroid, *P* is the intersection of the medians of the
triangle. Below is a GSP sketch of
this case.

Clearly, the ratio
holds in this case.

Case2:
*P* is the orthocenter. By definition of orthocenter, *P* is the intersection of the lines of the altitudes of
the triangle. Below is a GSP
sketch of this case.

Clearly, the ratio
holds in this case.

Case3:
*P* is the incenter. By definition of incenter, *P* is the intersection of the angle bisectors of the
triangle. Below is a GSP sketch of
this case.

Clearly, the ratio
holds in this case.

Case4:
*P* is the circumcenter. By definition of circumcenter, *P* is the intersection of the perpendicular bisectors of
the sides of the triangle. Below
is a GSP sketch of this case.

Clearly, the ratio
holds in this case.

Proof: Now I will show that given a triangle,
the three lines,,
and
intersect at point *iff
*.

First
I will assume that the three lines intersect at point *P* and then I will show that.

Below
is the triangle.

To
aid me in my proof, I will extend the segments *AD*, *BE*, and
*CF* to be lines and I will draw a
line through *C*, which is parallel
to *AB*. Below is the new sketch.

Now
I will shade in a couple of the similar triangles that occur.

Triangle
*GCP* ~ Triangle *AFP* because

1)
by congruency of alternate interior angles

2)
by congruency of vertical angle theorem

and 3) since and ,

and
by 1) and 2)

so
subtracting <*PGC* and <*CPG* from both sides yields .

Similarly,
Triangle *GCD* ~ Triangle *ABD*, Triangle *HCP* ~ Triangle *BFP* and
Triangle *HEC* ~ Triangle *BEA*,

* *Since the preceding triangles are similar, the
following ratios occur:

(1)

(2)

(3)

(4)

To
show that,
I can show that.

By
using (1), (3) and the property of transitivity, I get that . Solving for *HC*, I get . Solving for *HC* in (4) yields . By transitivity, . Now I will cross-multiply and get . Then, solving for *GC* in (2) yields and when I substitute that into my
equality, I get . By simplifying and cross-multiplying, I
get . Hence, .

Thus
far, I have shown one part of the proof. Now I will show the other part of the
proof. I must assume that and then show that *P* must be a point of concurrence.

Below
is a GSP sketch of the triangle that is given.

Assume
that *BE* and *CF* intersect at *P* and *AX *has been
constructed to go through *P*, but.

Since
*AD* and *AX* both go through *P*, we know that and .

Then,
by transitivity, .

After
simplifying, I get .

So .

Hence,
*P* is the point of concurrence of ,,
and. Therefore, I have proved the
theorem.

A
question that arises is how might this theorem help us to prove concurrency of
the lines of altitudes if *P* is the orthocenter?

First
we need a sketch of when *P* is the orthocenter.

The
theorem states that given a triangle,
the three lines,,
and
intersect at point *iff
*.

Let
*P *be the orthocenter. Then, by definition of orthocenter, the
lines through *P* from each vertex
must be perpendicular to the sides of the triangle. Notice that Triangle *ABD* is similar to Triangle *CBF* by angle-angle similarity (they share the angle *B* and they each have a right angle by
construction). Similarly (ha ha!),
Triangle *ACD* is similar to
Triangle *BCE* and Triangle *BAE* is similar to Triangle *CAF*. So the
following proportions result:

, and

By
CevaŐs Theorem, the three lines,,
and
intersect at point *iff
*. LetŐs
assume that . Substituting values from above, gives
us . Therefore, the
altitudes must be concurrent.

Next I will
generalize my result so that the point *P* may be outside the triangle.
In order to do that, I must extend the sides of the triangle to be
lines. A working sketch of this
case may be found here.