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.

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.

cot(w) = (a

m

and expressing them in terms of the sum of the squares of the sides we find that:

m

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.

area(FGH) = (1/3)(m

where (m

Let now k be the similarity ratio of FGH to the median triangle BDE, so that

k

Introducing this into (1) we obtain

area(FGH) = (1/3)(m

k = (1/3)(m

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)*cot

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 (a

S = area(FGH) - area(ABC) = ((4/3)*cot

area(DEF)/area(ABC) = (4/3)*cot

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.

Brocard.html

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