The Dual Theorem concerning Aubert Line

Professor Ion Patrascu, National College "Buzeşti Brothers" Craiova - Romania Professor Florentin Smarandache, University of New Mexico, Gallup, USA

In this article we introduce the concept of Bobillier transversal of a triangle with respect to a point in its plan; we prove the Aubert Theorem about the collinearity of the orthocenters in the triangles determined by the sides and the diagonals of a complete quadrilateral, and we obtain the Dual Theorem of this Theorem.

Theorem 1 (E. Bobillier)

Let 퐴퐵퐶 be a triangle and 푀 a point in the plane of the triangle so that the perpendiculars taken in 푀, and 푀퐴, 푀퐵, 푀퐶 respectively, intersect the sides 퐵퐶, 퐶퐴 and 퐴퐵 at 퐴푚, 퐵푚 and 퐶푚. Then the points 퐴푚, 퐵푚 and 퐶푚 are collinear.

퐴푚퐵 Proof We note that = 퐴푚퐶 aria (퐵푀퐴푚) (see Fig. 1). aria (퐶푀퐴푚)

1 Area (퐵푀퐴푚) = ∙ 퐵푀 ∙ 푀퐴푚 ∙ 2 sin(퐵푀퐴푚̂ ).

1 Area (퐶푀퐴푚) = ∙ 퐶푀 ∙ 푀퐴푚 ∙ 2 sin(퐶푀퐴푚̂ ).

Since

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3휋 푚(퐶푀퐴푚̂ ) = − 푚(퐴푀퐶̂ ), 2 it explains that

sin(퐶푀퐴푚̂ ) = − cos(퐴푀퐶̂ );

휋 sin(퐵푀퐴푚̂ ) = sin (퐴푀퐵̂ − ) = − cos(퐴푀퐵̂ ). 2

Therefore:

퐴푚퐵 푀퐵 ∙ cos(퐴푀퐵̂ ) = (1). 퐴푚퐶 푀퐶 ∙ cos(퐴푀퐶̂ )

In the same way, we find that:

퐵푚퐶 푀퐶 cos(퐵푀퐶̂ ) = ∙ (2); 퐵푚퐴 푀퐴 cos(퐴푀퐵̂ )

퐶푚퐴 푀퐴 cos(퐴푀퐶̂ ) = ∙ (3). 퐶푚퐵 푀퐵 cos(퐵푀퐶̂ )

The relations (1), (2), (3), and the reciprocal Theorem of Menelaus lead to the collinearity of points 퐴푚, 퐵푚, 퐶푚.

Note Bobillier's Theorem can be obtained – by converting the duality with respect to a circle – from the theorem relative to the concurrency of the heights of a triangle.

Definition 1 It is called Bobillier transversal of a triangle 퐴퐵퐶 with respect to the point 푀 the line containing the intersections of the perpendiculars taken in 푀 on 퐴푀, 퐵푀, and 퐶푀 respectively, with sides 퐵퐶, CA and 퐴퐵.

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Note The Bobillier transversal is not defined for any point 푀 in the plane of the triangle 퐴퐵퐶, for example, where 푀 is one of the vertices or the orthocenter 퐻 of the triangle.

Definition 2 If 퐴퐵퐶퐷 is a convex quadrilateral and 퐸, 퐹 are the intersections of the lines 퐴퐵 and 퐶퐷, 퐵퐶 and 퐴퐷 respectively, we say that the figure 퐴퐵퐶퐷퐸퐹 is a complete quadrilateral. The complete quadrilateral sides are 퐴퐵, 퐵퐶, 퐶퐷, 퐷퐴, and 퐴퐶, 퐵퐷 and 퐸퐹 are diagonals.

Theorem 2 (Newton-Gauss)

The diagonals’ means of a complete quadrilateral are three collinear points. To prove Theorem 2, refer to [1].

Note It is called Newton-Gauss Line of a quadrilateral the line to which the diagonals’ means of a complete quadrilateral belong.

Teorema 3 (Aubert)

If 퐴퐵퐶퐷퐸퐹 is a complete quadrilateral, then the orthocenters

퐻1, 퐻2, 퐻3, 퐻4 of the triangles 퐴퐵퐹, 퐴퐸퐷, 퐵퐶퐸, and 퐶퐷퐹 respectively, are collinear points.

Proof Let 퐴1, 퐵1, 퐹1 be the feet of the heights of the triangle 퐴퐵퐹 and 퐻1 its orthocenter (see Fig. 2). Considering the power of the point 퐻1 relative to the circle circumscribed to the triangle ABF, and given the triangle orthocenter’s property according to which its symmetrics to the triangle sides belong to the circumscribed circle, we find that:

퐻1퐴 ∙ 퐻1퐴1 = 퐻1퐵 ∙ 퐻1퐵1 = 퐻1퐹 ∙ 퐻1퐹1.

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This relationship shows that the orthocenter 퐻1 has equal power with respect to the circles of diameters [퐴퐶], [퐵퐷], [퐸퐹]. As well, we establish that the orthocenters 퐻2, 퐻3, 퐻4 have equal powers to these circles. Since the circles of diameters [퐴퐶], [퐵퐷], [퐸퐹] have collinear centers (belonging to the Newton-Gauss line of the 퐴퐵퐶퐷퐸퐹 quadrilateral), it follows that the points

퐻1, 퐻2, 퐻3, 퐻4 belong to the radical axis of the circles, and they are, therefore, collinear points.

Notes

1. It is called the Aubert Line or the line of the complete quadrilateral’s orthocenters the line to which the orthocenters 퐻1, 퐻2, 퐻3, 퐻4 belong.

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2. The Aubert Line is perpendicular on the Newton-Gauss line of the quadrilateral (radical axis of two circles is perpendicular to their centers’ line).

Theorem 4 (The Dual Theorem of the Theorem 3)

If 퐴퐵퐶퐷 is a convex quadrilateral and 푀 is a point in its plane for which there are the Bobillier transversals of triangles 퐴퐵퐶, 퐵퐶퐷, 퐶퐷퐴 and 퐷퐴퐵; thereupon these transversals are concurrent.

Proof Let us transform the configuration in Fig. 2, by duality with respect to a circle of center 푀.

By the considered duality, the lines 푎, 푏, 푐, 푑, 푒 and 푓 correspond to the points 퐴, 퐵, 퐶, 퐷, 퐸, 퐹 (their polars).

It is known that polars of collinear points are concurrent lines, therefore we have: 푎 ∩ 푏 ∩ 푒 = {퐴′}, 푏 ∩ 푐 ∩ 푓 = {퐵′}, 푐 ∩ 푑 ∩ 푒 = {퐶′}, 푑 ∩ 푓 ∩ 푎 = {퐷′}, 푎 ∩ 푐 = {퐸′}, 푏 ∩ 푑 = {퐹′}.

Consequently, by applicable duality, the points 퐴′, 퐵′, 퐶′, 퐷′, 퐸′ and 퐹′ correspond to the straight lines 퐴퐵, 퐵퐶, 퐶퐷, 퐷퐴, 퐴퐶, 퐵퐷.

To the orthocenter 퐻1 of the triangle 퐴퐸퐷, it corresponds, by duality, its ′ ′ ′ polar, which we denote 퐴1 – 퐵1 − 퐶1, and which is the Bobillier transversal of the triangle 퐴’퐶’퐷’ in relation to the point 푀. Indeed, the point 퐶’ corresponds to the line 퐸퐷 by duality; to the height from 퐴 of the triangle 퐴퐸퐷, also by ′ duality, it correspond its pole, which is the point 퐶1 located on 퐴’퐷’ such that ̂ ′ 0 푚(퐶′푀퐶1) = 90 .

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′ To the height from 퐸 of the triangle 퐴퐸퐷, it corresponds the point 퐵1 ∈ ̂ ′ 0 퐴′퐶′ such that 푚(퐷′푀퐵1) = 90 .

′ Also, to the height from 퐷, it corresponds 퐴1 ∈ 퐶′퐷’ on C such that ′ ′ 0 푚(퐴 푀퐴1) = 90 . To the orthocenter 퐻2 of the triangle 퐴퐵퐹, it will ′ ′ ′ correspond, by applicable duality, the Bobillier transversal 퐴2 − 퐵2 − 퐶2 in ′ ′ ′ the triangle 퐴 퐵 퐷 relative to the point 푀. To the orthocenter 퐻3 of the ′ triangle 퐵퐶퐸, it will correspond the Bobillier transversal 퐴3′ − 퐵 3 − 퐶3′ in the ′ ′ ′ triangle 퐴 퐵 퐶 relative to the point 푀, and to the orthocenter 퐻4 of the ′ ′ ′ triangle 퐶퐷퐹, it will correspond the transversal 퐴4 − 퐵4 − 퐶4 in the triangle 퐶′퐷′퐵′ relative to the point 푀.

′ ′ ′ ̅̅̅̅ The Bobillier transversals 퐴푖 − 퐵푖 − 퐶푖 , 푖 = 1,4 correspond to the collinear points 퐻푖, 푖 = 1̅̅,̅4̅.

These transversals are concurrent in the pole of the line of the orthocenters towards the considered duality.

It results that, given the quadrilateral 퐴’퐵’퐶’퐷’, the Bobillier transversals of the triangles 퐴’퐶’퐷’, 퐴’퐵’퐷’, 퐴’퐵’퐶’ and 퐶’퐷’퐵’ relative to the point 푀 are concurrent.

References

[1] Florentin Smarandache, Ion Patrascu: „The Geometry of Homological Triangles”. The Education Publisher Inc., Columbus, Ohio, USA – 2012.

[2] Ion Patrascu, Florentin Smarandache: „Variance on Topics of Plane Geometry”. The Education Publisher Inc., Columbus, Ohio, USA – 2013.

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