The median triangle of triangle ABC is the triangle created by drawing successive parallel and equal segments to the medians of ABC starting from an arbitrary point.
Below I start from B and the median BD and draw DE, EB parallel to the other medians of the triangle.
The following properties are easy consequences of the definition.
[1] The medians of the median triangle are parallel and equal to 3/4 of corresponding sides of the original triangle.
[2] The area of the median triangle is also 3/4 of the area of the original triangle.
[3] The median of the median triangle is homothetic to the original triangle.
[4] The median triangle is similar to triangle FGH formed by the orthogonals to the medians from the vertices of ABC. For the similarity ratio see the fourth section below.
Given the triangle ABC, the Vecten configuration of the triangle is the figure resulting by constructing squares on the sides of the triangle (this is studied in Vecten.html ). It is proved that the so called flanks which are the triangles {ADI, GHC, EBF} have their external sides equal to the double and orthogonal to the respective medians of the triangle (e.g. GH is double and orthogonal to the median MC). Besides they have also the same area with the triangle of reference ABC.
Consequently, if put together these flanks (like in the triangle ED-FG-HI above) make a triangle similar to the median triangle. This triangle composed from flanks is similar to the triangle DEF of the figure below, created by extending the exterior sides of the flanks. Because of property 1.3 repeating the construction of the flanks and taking again the exterior sides creates triangle GHI (below) which is similar to the original.
In fact, the properties contained in the first section imply that the two triangles {ABC, DEF} are homothetic. Their homothety ratio is calculated in the last section below.
The Brocard angle of a triangle is discussed in Brocard.html . This angle (w) satisfies the equation
cot(w) = (a2+b2+c2)/(4D), D denoting the area of triangle ABC. Thus, the Brocard angle relates naturally to the Vecten configuration since cot(w) is expressed as the quotient of the sum (of areas) of its squares to the sum of its triangles. Replacing in this formula the squares with the squares of the medians, which are given by:
ma2 = (1/4)(b2 + c22bccos(A)) and its cyclic permutations of letters,
and expressing them in terms of the sum of the squares of the sides we find that:
ma2 + mb2 + mc2 = (3/4)(a2 + b2 + c2)
and consequently (see also 1.2) the Brocard angle of the median triangle coincides with the corresponding of the triangle of reference. Such triangles are called equi-Brocardian.
The similarity ratios of the various cases discussed above involve all the Brocard angle of the triangle of reference. To start with consider the area of the triangle FGH of the first section. It is composed from areas of triangles {IFG, IGH, IHF}. The area of a typical of them is (1/2)GH*IC = (1/3)mc*c', hence
area(FGH) = (1/3)(ma*a' + mb*b' + mc*c'), (1)
where (mi) denote the medians of ABC and {a',b',c'} the corresponding orthogonal to them sides of FGH.
Let now k be the similarity ratio of FGH to the median triangle BDE, so that
k2 = area(FGH)/area(BDE), and a'=kma, b'=kmb, c'=kmc. (2)
Introducing this into (1) we obtain
area(FGH) = (1/3)(ma2 + mb2 + mc2)*k = k2* area(BDE) =>
k = (1/3)(ma2 + mb2 + mc2)/area(BDE) =>
k = (4/3)cot(w), (3)
where (w) the Brocard angle of the triangle of reference ABC (same with Brocard angle of BDE).
From this and the fact that area(ABC) = (4/3)area(BDE) follows that
area(FGH)/area(ABC) = (4/3)*cot2(w). (4)
From this we can calculate the area of triangle DEF of the last figure and especially its ratio to the area of the triangle of reference ABC. In fact triangle DEF has an area composed of the three squares (a2 + b2 + c2), four times the area(ABC) and the sum (S) of the remaining three triangles which when shifted paralllel to themselves have a side coincident with one side of ABC build up triangle FGH of the first section minus the area of ABC. Thus
S = area(FGH) - area(ABC) = ((4/3)*cot2(w)-1)*area(ABC), and from this
area(DEF)/area(ABC) = (4/3)*cot2(w) + 4*cot(w) + 3 (5).
Since GHI in the last figure is constructed from DEF in the same way later is constructed from ABC, if follows that the same number gives the homothety ratio of GHI to ABC.
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