The Department of Mathematics
Education
William Daly
EMAT6680
Final Write Up
Summer
03
Relationships
Among Segmented Sides of a Triangle
Shown below is an arbitrary acute triangle with an
arbitrary interior point. The sides are
constructed of extended lines which are hidden for clarity. The construction was done on the extended
lines to allow for exploration of the point P outside of the triangle.
Segments from a vertex to the arbitrary point is
shown. A relationship appears to emerge
between the product of the lengths of the three side segments shown in bold
(AF, BD and CE) and the product of those shown in dashed lines (FB, DC, and
EA). The relationship is that these
products appear to be equal as shown in the measurements on the figure
above. Although this appears to be a
clear relationship, that is yet to be proved.
A hint was given to consider parallel lines and
similar triangles as a strategy for formulating a proof. To engage this hint, start with a line
parallel to the base of the triangle through the vertex A:
Notice
that one would like to establish relationships among the segments within this
triangle associated with AF, FB, BD, DC, CE and EA. So it would appear to be beneficial to unhide the extended lines through
point P in search of these similarities.
A
number of similarities containing the segments of interest are readily
observable. From knowing the equality
of interior angles of a line intersecting parallel lines and from knowing the
equality of angles opposite the intersection of two line, the following
observations are made:
Noting
these equality of these pairs of angles, several pair of similar triangles can
be identified. To find relationships among
the various segment, one may suppose that including a segment of interest along
with a segment common to other similar triangle into a set of similar triangles
may be useful. For example, we would
like to say something about the segment AF and AE, so note relationships among
the following pair of similar triangle (the green area is just where the blue
triangle BFC is overlapping triangle BEC):
From
this figure the following relationships between similar triangles may be
stated:
and 
In
looking at these triangles we have included a statement about every segment
except BD and DC, and we have said nothing about segments common between pairs
of similar triangle that may be used to relate these pairs to each other. To attend to this, additional similar
triangles are observed that seem to meet this need:
The
following relationship may be stated:
and 
Note
that the segments PD and AP provide a link between the two relationships
between similar triangles, so that all of these ratios are equal:
Now
summarize the relationships we have discovered so far.
(1) 
(2) 
(3) 
Inspect
these three equations for the segments of interest. The first term of equations (1) and (2) have four of the six
segments in question (AF, BF, CE, and AE) so obviously we would like to keep
these. The only other two are CD and BD
which appear in the first and fourth terms of equation (3). The encouraging news is that if I decide to
eliminate all but the first and fourth terms of equation (3), there are lengths
AG which will link to equation (2) and AH which will link to equation (1). Let’s just hope that the length BC
disappears. So rewriting equations (1)
through (3) based on the above discussion, we are left with:
(1) 
or 
(2) 
or 
(3) 
or 
Substitute
(1) and (2) into (3):
As
we had hoped, the length BC may be cancelled from both side of the
equation. The result becomes:
or 
, which, looking back at the original figure, is the result
we are trying to prove.
A question arises as to whether this relationship
holds for points P outside of the triangle.
Click hereto see an Animation of P
moving along a path inside and outside of the triangle. The demonstration shows that the
relationship between the products, shown side by side, appears to hold
regardless of the location of P.
A further investigation explores the relationship of
the triangles ABC and DEF, where point P is within the boundary formed by
ABC. It is queried whether the ratio of
the areas of ABC to DEF, with point P constrained in this way, is always
greater than or equal to 4. Where is
equal to 4? This is not proven here,
but explored by demonstration. Clear
here to see the Demonstration. As shown, the ratio of the areas is always
greater than 4, except at a single point.
This point has the appearance of being the Centroid. This is empirically confirmed by clicking
the action button in the demonstration.