Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, CP and BP extended to their intersections with the opposite sides in points D, E and F, respectively.
This investigation will explore the line segments of the sides of ABC which have been generated by the lines going through P. We will make a conjecture about their relationship and prove it by using, among other things, the sub-triangles formed by the lines.
First we will explore line segments created by the extension of AP, CP and BP to D, E and F, respectively.† By measuring the lengths of the segments, we notice that (AF)*(DC)*(BE) = (FC)*(DB)*(EA) for any location of P inside ABC, and regardless of whether ABC is acute or obtuse. By the same token, the ratio of (AF)*(DC)*(BE) to (FB)*(DB)*(EA) is always one.
We begin our proof by constructing a line parallel to BC, going through A. We then extend the lines CE and BF such that they intersect this parallel line at G and H, respectively.
By constructing this parallel line, we have generated four pairs of similar triangles. Two pairs each are outlined in the first and second diagrams that follow.
With similar triangles we know that corresponding sides are proportional to each other. The proportional relationship between each pair of similar triangles is indicated in the captions included with the diagrams above.
We are most interested in the identities that involve the segments of the sides of triangle ABC, and so throughout our analysis they have been emphasized and color-coded to match the diagrams.
Using algebra, we multiply the left sides of the identities stemming from the similar triangles, and set them equal to the product of the right sides of the same identities.† That is,
(AF/FC)(BE/EA)(DC/DB) = (AH/CB)(BC/AG)(AG/AH)
By taking specific measurements listed below, we confirm that both sides of this equation are equal to each other, and that their value is equal to one.
When P is inside triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to four. Notice that it is exactly equal to four when D, E and F are the midpoints of their respective sides, and therefore P is the centroid of triangle ABC.† CLICK HERE for a GSP demonstration of this.
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