It's Not "Napoleon's Theorem"...But It's Close!

by : Dixie Williford

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.

Then, 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.

Proof:

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.

Segment 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.

Similarly, 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.

As previously 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, ÐWEF = 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. q.e.d.

II. Now 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 WD.)

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 following information:

WE + ED = AB + BT

ED/WE = r = JE/KE

BT/AB = s = BH/GB

We want to obtain the equation:

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:

KE + r(KE) = GB + s(GB).

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) = GB(1+s)

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.