Theorem: Let ABCDEF
be any arbitrary hexagon such that upon each side of ABCDEF there
are constructed triangles: ASB, BTC, CUD, DVE, EWF and FXA. Each
triangle is formed by extending the adjacent sides of a given
segment until the two adjacent sides intersect. Let the centroids
of these six triangles be: G, H, I, J, K, L, respectively.
the hexagon GHIJKL has opposite sides congruent and parallel.
Furthermore, the diagonals, of GHIJKL are concurrent at a point,
O, that is the midpoint of each diagonal.
I. We will begin
by proving that the opposite sides of GHIJKL are parallel. Let
us look at the two triangles: FXA and ASB. By definition of the
hexagon ABCDEF, we know that segments XF and SB are parallel.
SF serves, then, as a transversal. Therefore, angle SFA is congruent
to angle ABS. Similarly, segment XB serves as a second transversal
to XF and SB and, therefore, angle FXA is congruent to angle SBA.
We also have that angle XAF is congruent to angle SAB because
they are vertical angles. Therefore, by the AAA similarity criterion,
triangle FXA is similar to triangle ASB. (By repeating this argument
for every remaining possible combination of triangles among: ASB,
BTC, CUD, DVE, EWf, FXA, each pair of triangles is similar by
the AAA criterion.)The following triangles have been removed from
the original configuruation so that we can focus our attention
on segments KJ an dGH (which are opposite sides in the hexagon
GHJIKL) and prove that they are parallel to one another. In triangle
WFE, let point N be the midpoint of side FE, P be the midpoint
of side WE, and M be the midpoint of side WF. Therefore, WN FP
and EM are the three medians of triangle and interesect at the
centroid (by definition of centroid), point K.
in triangle TBC, let R be the midpoint of side TC, Z be the midpoint
of side Bc and Q be the midont of side Bt. Therefore, segments
BR, TZ and CQ are the three medians of the triangle and they interesect
tat the centroid (by definition of centroid), point H.
proven, triangles TBC and WFE are similar triangles. Therefore,
corresponding segments are proportional and coreresonding angles
are congruent. We know from the definition of our original hexagon,
ABCDEF, that FE is parallel to BC. We, therefore, have the following
ratios established: ,
for some positive integer a. Also, the following angles are congruent:
Ð WFE = ÐTCB,
= ÐTBC, ÐCTB
= ÐEWF. Similarly,
every other pair of corresonding segments are proportional to
one another with ratio, a, and every other pair of corresonding
angles are congruent (this is by the definition of similar triangles).
Well, K is the centroid
of triangle TBC. Also, n is the midpoint o FE and Z is the midpoint
of BC, two corresponding segments. Furthermore, point B corresonds
to point E (B is the oint of intersection of TB and CB, E is the
point of intersection of WE and FE, and WE corresponds to TB and
RE corresponds to HB, KN corresponds to HZ and FN corresponds
to BZ with the ratio between corresponding to segments equivalent
to a. Therefore, triangle HBZ is similar to triangle KEN. Therefore
angle HBZ is congruent to angle KEN by definiton of similar triangles.
We know from the definition of our original hexagon, ABCDEF, that
segment FE is parallel to segment BC (which are both opposite
sides of ABCDEF). Therefore segment KE makes an angle, KEN, with
FE which is congruent to the Ð HBZ that segment HB makes with
segment BC ( with BC parallel to FE). Therefore, the line KE is
parallel to the line HB. Well, point J (the centroid of triangle
JED, adjacent to triangle WFE) lies on the line KE. similarly,
point G (the centroid of triangle SBA, adjacent to triangle TBC)
lies on the line HB. Therefore, segment KJ (a subset of line KE)
is parallel to segment HG ( a subset of line HB). By repeating
this proof for the remaining two pair of opposite sides of GHIJKL,
we have that each pair of opposite sides of GHIJKL are parallel.
I will prove that each pair of opposite sides of GHIJKL are congruent.
Again, I will first show that the two opposite sides, segment
KJ and segment GH are congruent. For this portion of the proof,
we will be focusing in on the parallelogram WDTA. (WDTA is forced
to be a parallelogram by the intersection of two sets of parallel
lines: line AW parallel to line Td, and line ATparallel to line
We will now use that
WD = AT in order to show that KJ = GH. Again, let us remember
that triangle WFE is similar to triangle VEd. Therefore, for every
pair of corresponding segments between the two triangles, we can
establish a ratio equivalent to r, for some positive integer r.
Likewise, triangle SAB
is similar to triangle TBC. Therefore, for every pair of corresponding
segments between the two tirangles, we can establish a ratio equivalent
to s, for some positive integer, s.
Therefore, we know the
WE + ED = AB + BT
ED/WE = r = JE/KE
BT/AB = s = BH/GB
We want to obtain the
KE + EJ = GB + BH.
[Remember that JE/KE = r and BH/GB = s. Therefore, JE = r(KE)
and BH = s(GB).
Therefore, it is equivalent
to say that we want to obtain the equation:
By showing that our
given information leads to this latter equation, we will have
shown that opposite sids of GHIJKL are congruent.
Manipulating the equations
that the relationships that we are given, we obtain the following:
ED = r(WE). Therefore,
WE + ED = r(WE) + WE
BT = s(AB). Therefore,
AB + BT = s(AB) + AB
Therefore, WE + ED =
AB + BT yields: r(WE) + WE = s(AB) + AB.
Therefore, we obtain
*WE (1+r) = AB(1+s).
It is now convenient
to introduce a new piece of information. As proven earlier, each
pair of the six triangles formed on the sides of ABCDEF are similar
to one another. Therefore, we have the ratio:
**KE/WE = GB/AB
Therefore, we can multiply
equation(*) through by our newly obtained ratio (**).
(KE/WE)[WE(1+r)] = (GB/AB)[AB(1+s)]
Therefore, KE(1+r) =
Therefore, KE + r(KE)
= GB + s(GB). So, KJ = GH. By repeating this same argument for
each pair of opposite sides of the hexagon GHIJKL, we have that
each pair of opposite sides of GHIJKL are congruent.
III. It is now easy to show that
the diagonals of the hexagon GHIJKL are concurrent at the point
O, whichis also the midpoint of each diagonal. Let us first show
this property for the two diagonals, segment LI (connecting opposite
vertices Land I of GHIJKL) and segment GJ (connecting opposite
vertices G and J of GHIJKL).
Because we know that
the sides of GHIJKL are parallel and congruent, we can construct
the parallelogram GLJI (segment GL and segment JI are parallel
and opposite one another in the hexagon GHIJKL). Therefore, the
diagonals of GLJI are segment LI and segment GJ (also two diagonals
of GHIJKL). By property of parallelograms, LI and GJ(also two
diagonals of GHIJKL). By property of parallelograms, LI and GJ
intersect at O, such that O is the midpoint of both LI and GJ.
Now let us look at a
second pair of diagonals of hexagon GHIJKL. Let us lookat the
pair segment LI and segment KH. Once again, because we know that
segment LK and segment HI are parallel and congruent, we can construct
the parallelogram LKIH, such that segment LI and segment KH are
also the diagonals of LKIH. Therefore, we know, by property of
parallelograms, that LI and KH intersect precisely at the midpoint
of each segment. Well, a segment only has one midpoint (by definition
of midpoint). We discovered earlier that the midpoint of segment
LI is the point O. This point O must also be the midpoint of segment
KH (by property of parallelograms). This is the same point O that
was the point of intersection of segment LI and segment GH. Therefore,
segment GJ, segment LI and segment KH are concurrent at the point
O, which is the midpoint of each of the three segments. Similar
construction and argument for each of the other pairs of diagonals
of GHIJKL show that each pair of diagonals of hexagon GHIJKL intersect
precisely at the point O, which is the midpoint of each of the
diagonals of hexagon GHIJKL.