Sinje J. Butler

The following triangle has been constructed by considering
any triangle **ABC** and selecting a
point **P** inside the triangle. Lines **AP,
BP****, **and **CP** were drawn and extended to their intersections
with the opposite sides in point **D, E **and
**F** respectively.

Please click here for a **GSP
Demonstration **exploring **(AF)(BD)(CE)** and **(FB)(DC)(EA) **for various triangles and various locations
of **P.
**

** **

The GSP Demonstration suggests that for all triangles and
all points of **P**, **(AF)(BD)(CE) = (FB)(DC)(EA). **Is this true? Even though it appears to be the case,
the GSP Demonstration does not prove it to be true. Below is a proof of 1 case where this is true. The other cases are left for other
explorations.

** **

Let’s begin by proving

**(AF)(BD)(CE) = (FB)(DC)(EA)**

** **

for the case of triangle **ABC**
at point **P **that is pictured above.
We will prove by constructing some parallel lines that produce similar
triangles and using the related ratios. *In coming up with a proof, the similar triangles
that I constructed did not provide any segments that allowed for
substitutions. Therefore, I
used, www.cut-the-knot.org/Generalization/ceva.shtml
for a hint on how to construct the correct similar triangles**.*

In the picture below, a line through point **A** parallel to segment **BC** was constructed. Segments **CP****
**and **BP**
were extended to lines and there intersections with the parallel line through
point **A** were marked as **G** and **H**
respectively. Within this
construction lie similar triangles that will be used for the proof.

Angle **GPA** is
equal to angle **DPC** by vertical
angles. Also, because angle **AGP** and angle **PCD** are alternate interior angles, they are equal. So, by the AA Similarity Postulate

triangle **GPA** is
similar to triangle **DPC**.
Therefore, we know the following

.

Also, from the above picture we can use the same line of
arguing to show that triangle **AHP**
is similar to triangle **PBD**. Angle **HPA**
is equal to angle **BPD **by vertical angles. Also, because angle **AHP** and angle **PBD** are alternate interior angles, they are equal. So, by the AA Similarity Postulate
triangle **AHP **is similar to
triangle** BPD**.
Therefore, we know the following

The ratio of the corresponding sides of triangle **GPA **and triangle **DPC** is equal to

the ratio of the corresponding sides of triangle **AHP **and triangle **BPD**. From
this we can write the following

.

Because

are line segments that we are not interested in, we can eliminate them and write the following for future use.

From the picture below, we can argue that triangle **AHE** is similar to triangle **BEC**.

Angle **AEH** is
equal to angle **BEC** by vertical
angles. Also, because angle **AHE** and angle **EBC** are alternate interior angles, they are equal. So, by the AA Similarity Postulate

triangle **AHE **is
similar to triangle **BEC**.
Therefore, we know the following

Because

contains segments that we are not interested in, we can eliminate it and write the following for future use.

.

Triangle **GAF** is
similar to triangle **FBC.**

Angle **GFA** is
equal to angle **BFC** by vertical
angles. Also, because angle **AGF** and angle **FCB** are alternate interior angles, they are equal. So, by the AA Similarity Postulate

triangle **GAF **is
similar to triangle **BFC**.
Therefore, we know the following

.

Because

contains segments that we are not interested in, we can write the following for future use.

Recall the three sets of ratios that we established for future use.

**(1) **,

**(2) **,

and

**(3) **.

Number three can be rewritten as

.

Number two can be rewritten as

.

Plugging in number two and three into number one gives the following.

Which can also be written as

or

.

Divide both sides by **BC, **which
gives

or

** **

Click here for a **GSP Demonstration
**that suggests when **P** is inside triangle **ABC**, the ratio of the areas of triangle **ABC** and triangle **DEF** is always greater than or equal to **4**. And it also appears that the ratio is
equal to four when triangle **DEF** is
the medial triangle. This is not a
proof, it can be used for further exploration.