Janet Kaplan

Final Assignment

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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.

CLICK HERE for a GSP file in
which you can move point P inside ABC yourself. Notice that the product of the
red line segments always equals the product of the green line segments.

Actually, this is a theorem, known as Ceva’s Theorem. It defines the segments through P from each
vertex to the opposite sides of the triangle as Cevians and states: Given any triangle ABC
with a point P in the interior, the product of the ratios of the pairs of
segments formed on each side of the triangle by the intersection point is equal
to one, where the ratios are taken in the same orientation on each side.
Further, if the ratio formed by any three Cevians is
equal to one, then the three Cevians are concurrent.

Stated algebraically, (AF/FC)(BE/EA)(DC/DB) = 1 or (AF)(BE)(DC) / (FC)(EA)(DB) = 1

Notice that this is the case with our triangle ABC and the
arbitrary interior point P.

Unlike many relationships in Geometry, this result is not
intuitive to us. Let us prove this conjecture and Ceva’s
Theorem.

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.

Since the value of (AF/FC)(BE/EA)(DC/DB) = 1, the values of (AF*BE*DC)
and (FC*EA*DB) are equal to each other.

This proves our original conjecture: (AF)*(DC)*(BE) = (FC)*(DB)*(EA)

** I
received help on this proof from www.cut-the-knot.org/Generalization/ceva.shtml

Can this result be generalized so
that point P can be outside the triangle? See for yourself by CLICKING HERE.

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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