[alogo] Vecten Configuration (5)

On the sides of triangle t=(ABC) erect squares.
The triangles ADI, CHG, EBF are known as [flanks].

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]
[2_0] [2_1] [2_2]

Erect the square BCLN on the other side of BC. Triangle t'=(CHL) is equal to t. They have equal sides (LC=BC and CH=AC) and equal angles at C. By [Vecten4], if O is the middle of GH, CO is orthogonal to AB, hence LH is equal and parallel to EB. Thus p=(HLEB) is a parallelogram. Now triangle t* = (MBC) is rotational-similar to t**=(EBL). Triangle s*=(KCJ) is also rotational similar to s**=(BCH). The similarity ratio being in both cases sqrt(2). Thus MC is equal to KJ and orthogonal to it.
As a result, the Vecten point is the circumcenter of the anticomplementary of MJK.

Look at the file: Vecten4.html for an other property of the above figure .

The file Vecten.html introduces the subject and discusses the first properties of the Vecten configuration of an arbitrary triangle.

The file Vecten_tiles.html introduces to the study of an interesting related subject.

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