**by**

**Behnaz
Rouhani**

The products appear to be equal for any triangle ABC and for any position of P. Below are two sketches for acute and obtuse triangles.

*Acute triangle*

*Obtuse triangle*

To explore this further with different types of triangles
and different positions of P click
here.

To prove that the products of the segments must be equal, we will draw some parallel lines to AD from C and B. Extend CF to intersect the parallel line at S and extend BE to intersect the other parallel line at R as shown below.

Now, we have several similar triangles. These four
pairs of similar triangles are important for the proof:

- Triangles BPD and BRC are similar by the angle-angle similarity theorem. In the given triangles (a) angle PBD is shared, and (b) angles BPD and BRC as well as angles BDP and BCR are congruent since PD is parallel to RC. From these triangles we could deduce:

- Triangles BSC and DPC are similar by the angle-angle similarity theorem. In these triangles (a) angle PCB is shared, and (b) angles CPD and CSB as well as angles CBS and CDP are congruent since PD is parallel to SB. From these triangles we could conclude:

- Next, we look at the similar triangles APE and REC. They are similar by the angle-angle theorem. In these triangles (a) angles AEP and REC are vertical angles, and (b) angles CRE and APE are alternate interior angles, which are congruent as AP is parallel to RC. Based on this similarity we could write:
- Finally, we consider triangles SBF and AFP, which are similar by the same reasoning as in the above case. This similarity would yield:

Multiplying the left-hand sides of equations (4),
(5), and (6) and multiplying the right-hand sides of the same equations
together would yield:

This is the proof of the first part of Ceva's Theorem which states that in a triangle ABC, three lines AD, CE, and BF intersect if and only if

This result can be generalized for points outside the triangle. As you can see in the sketch below this is the case.

To explore this further click here to view a Geometer's Sketchpad Sketch of the above. Move P around and observe the results.

When P is anywhere inside triangle ABC, the ratio
of the areas of triangle ABC and triangle DEF is always greater than 4.
The ratio, however, is equal to 4 when P is at the centroid of the triangle
ABC. To explore this further click
here and remember to use the move button to move P to point G which
is the centroid of triangle ABC. This is the only time ratio of areas will
be 4. Now let us prove this point.

Point P is the centroid of the triangle ABC, and
thus the lines AD, BE, and CF are the medians of triangle ABC. We know
that BD = DC,

CE = EA, and AF = FB. For simplicity we will use
the following notations:

**BD = DC = a ** or **BC
= 2a**

**CE = EA = b**
or **AC = 2b**

**AF = FB = c**
or
**AB = 2c**

Further, based on the Triangle Mid-segment Theorem we know that: In any triangle, a segment joining the midpoints of any two sides will be parallel to the third side and half its length. Thus,

**EF = a**

**DE = c**

**DF = b**

In order to find the area of triangle DEF we will use Heron's formula, which states that:

Where

Now, to find the area of triangle ABC we need to work on the semi-perimeter of triangle ABC

Using Heron's formula for area of triangle ABC would yield:

Or

**Area of Triangle ABC = 4 * Area of triangle DEF**

Based on the above, it can be concluded that when
P is at the centroid of triangle ABC the area of triangle ABC is four times
the area of triangle DEF.