A note on special cases of Van Aubel s theorem

Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 ISSN: 7-59 Journal homepage: www.ijaamm.com IJAAMM International Journal of Advances in Applied Mathematics and Mechanics A note on special cases of Van Aubel s theorem Dasari Naga Vijay Krishna Department of Mathematics Narayana Educational Instutions Machilipatnam Bengalore India Research Article Received 0 February 018; accepted in revised version 09 May 018 Abstract: MSC: In this short paper we study some properties of the lines associated with van Aubel s theorem in the special case when squares constructed on the sides of an arbitrary quadrilateral are replaced with equilateral triangles as well as isosceles triangles. 97K0 68R10 Keywords: Van Aubel s theorem Van Aubel s point Orthodiagonal quadrilateral Equilateral triangle Kiepert hyperbola 018 The Authors. This is an open access article under the CC BY-NC-ND license https://creativecommons.org/licenses/by-nc-nd/.0/. 1. Introduction Van Aubel s theorem Consider an arbitrary Quadrilateral ABCD the quadrilateral S 1 S S S formed by joining the four corresponding centers S 1 S S S of the squares thus constructed on each side of ABCD is an iso-ortho diagonal quadrilateral [6]. That is S 1 S S S and S 1 S S S. From Fig. 1 it is clear S 1 S S S and S 1 S S S. Fig. 1. E-mail address: vijay990009015@gmail.com 0

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 1 The theorem we just stated above is attributed to Van Aubel Von Aubel in [Gardner p. 176-178] could also be found in their work de Villiers Yaglom Finney among others. In this article we study the properties of the lines S 1 S and S S when the squares are replaced with equilateral triangles and isosceles triangles our present study about the special case of Van Aubel s theorem when the squares are replaced with equilateral triangles and further generalization is not actually new since some of the authors studied about this earlier in 90 s can be found in [ 5 7 10]. Even though it is not a new study and the results presented in this article seems to be very elementary but are quite new and interesting. In this short note we also study about a point named as Van Aubel s point its geometricalruler and compass construction its location in general case and few more generalizations of van Aubel s theorem associated with Kiepert hyperbola.. Preliminaries We use the following lemmas in proving the results. Lemma.1. If Ax 1 y 1 Bx y are the two vertices of an arbitrary triangle ABC whose base angles are A and B then the coordinates of third vertex C x y is given by x1 tan A + x tanb ± tan A tanb y 1 y tan A + tanb y1 tan A + y tanb tan A tanb x 1 x tan A + tanb x1 cotb + x cot A ± y 1 y cot A + cotb or y1 cotb + y cot A x 1 x cot A + cotb Proof. Consider x 1 y 1 1 Λcot A + cotb det x y 1 x 1 cotb + x cot A ± y 1 y y1 cotb + y cot A x 1 x cot A + cotb By doing row operation on R using R 1 and R we get x 1 y 1 1 Λcot A + cotb x y 1 ± y 1 y x1 x 0 Which implies {Λcot A + cotb} ± [x 1 x + ] y 1 y ±AB 0 We have area of triangle ABC 1 x 1 y 1 1 det x y 1 x 1 cotb+x cot A±y 1 y cot A+cotB y 1 cotb+y cot A x 1 x cot A+cotB 1 1 x 1 1 y 1 1 cot A + cotb det x y 1 x 1 cotb + x cot A ± y 1 y y1 cotb + y cot A x 1 x cot A + cotb 1 1 {Λcot A + cotb} cot A + cotb 1 1 cot A + cotb ±AB AB 0 since cot A + cotb 0 cot A + cotb It proves that area of triangle ABC is not equal to zero which means that there is a triangle with A B and C as vertices.

A note on special cases of Van Aubel s theorem Now let us prove that base angles of triangle ABC are A and B if the third vertex either C or C 1 where x1 cotb + x cot A + y 1 y C y 1 cotb + y cot A x 1 x cot A + cotb cot A + cotb and C 1 x1 cotb + x cot A y 1 y y 1 cotb + y cot A + x 1 x cot A + cotb cot A + cotb Clearly the midpoint D of C C 1 lies on the line AB since C 1 is the image of C with respect to the base AB of triangle ABC its coordinate is given by x1 cotb + x cot A D y 1 cotb + y cot A cot A + cotb cot A + cotb And also D divides AB in the ratio given by Hence AD DB cot A cotb Now AD AB cot A cot A + cotb DB AB cotb cot A + cotb C D C D AB AB cot A + cotb Since C DC D 1 are the heights of the triangle ABC triangle ABC 1 hence C D AD C D AD tan A C D DB C D DB tanb This proves that the base angles are A and B. Note:.1 is true even if one of the angles either A or B is right angle. Proof. We start with A 90 o So By considering the point C given in the.1 we have to prove that C Aand AB are perpendicular to each other and t anb C A AB. As we have angle A 90 0 so cot A 0 and cotb 0. Hence x1 cotb + y 1 y C y 1 cotb x 1 x cotb cotb Slope of the line C A Hence C A AB C A AB 1 AB y1 y cotb x1 x y 1 y Slope of the line AB x1 x + 1 AB tanb cotb AB cotb y1 y x 1 x. It is clear that slope of C A slope of AB 1 This Proves that the point C so defined as in the statement of the lemma is in fact the third vertex of the triangle ABC when A 90 0. A nalogously it is shown for C 1 the same occurs when B 90 0.

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 Corollary.1. If ang l e A ang l eb θ that is triangle ABC is an isosceles triangle then the coordinates of C are given by x1 + x ± tanθ y 1 y y1 + y tanθ x1 x Corollary.. If A B 60 o that is triangleabc is an equilateral triangle then the coordinates of C are given by x1 + x ± y 1 y y1 + y x1 x Lemma.. If Ax 1 y 1 Bx y C x y are the three vertices of an arbitrary triangle ABC then the coordinates of its circum center are given by x1 sina + x sinb + x sinc sina + sinb + sinc where ABC are the angles of the triangle. y 1 sina + y sinb + y sinc sina + sinb + sinc Corollary.. The coordinates of the circum center of an isosceles triangle whose vertices are Ax 1 y 1 Bx y x1 + x ± tanθ y 1 y y1 + y tanθ x1 x and C Where θ is the base angle are given by x1 + x cotθ y 1 y y1 + y ± cotθ x1 x. Lemma.. If x 1 y 1 x y x y are the three vertices of an equilateral triangle then the coordinates of its center are given by x1 + x + x y 1 + y + y. Corollary.. The coordinates of the center of an equilateral triangle whose vertices are x 1 y 1 x y and x1 +x ± y 1 y x1 x y1 y are given by + x 1 +x y 1+y x x 1 + y 1 +y. Main results Theorem.1. If S 1 S S S are the centers of the equilateral triangles ABP BCQ C DR D AT are constructed which lie entirely out wards on the sides AB abc bc D c and AD d of an arbitrary quadrilateral ABC D respectively then the lines PR QS are respectively perpendicular to the lines S S S 1 S. That is S 1 S QT S S PR. [] Proof. With out loss of generality let us consider the coordinates of vertices of the quadrilateral ABC D as A x 1 y 1 B x y C x y and D x y. Then using. we have x1 + x + y 1 y y1 + y x1 x P and x + x + y y Q y + y x x x + x + y y y + y x x R x + x 1 + y y 1 y + y 1 x x 1 T

A note on special cases of Van Aubel s theorem Fig.. From Fig. it is clear S 1 S QT and S S PR and it is clear that S 1 A + B + P y 1 y + x 1 + x B +C +Q y y + x + x S S C + D + R y y + x + x S D + A + T y y 1 + x + x 1 So x x 1 + y 1 + y x x + y + y x x + y + y x 1 x + y + y 1 y1 + y y y + x + x x x 1 Slope of the line PR x 1 + x x x y + y y y 1 y + y y y 1 + x + x x 1 x Slope of the line QT x + x x x 1 y + y y 1 y x 1 + x x x y + y y y 1 Slope of the line S S y1 + y y y + x + x x x 1 x + x x x 1 y + y y 1 y Slope of the line S 1 S y + y y y 1 + x + x x 1 x Now it is clear that slope of PR slope of S S 1 slope of QT slope of S 1 S. Hence S 1 S QT S S PR Theorem.. If S 1 S S and S are the centers of the equilateral triangles ABP BCQ C DR D AT are constructed which lie entirely inwards on the sides AB a BC b CD c and AD d of an arbitrary quadrilateral ABCD respectively then the lines P R Q T are respectively perpendicular to the liness S and S 1 S. That is S 1 S Q T S S P R.

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 5 Proof. Without loss of generality let us consider the coordinates of vertices of the quadrilateral ABC D as A x 1 y 1 B x y C x y and D x y. Then using. we have P Q R T x1 + x y 1 y x + x y y x + x y y x + x 1 y y 1 And it is clear that S B +C +Q y y 1 + x 1 + x 1 S C + D + R y y + x + x S C + D + R y y + x + x S D + A + T y 1 y + x + x 1 y1 + y + x1 x y + y + x x y + y + x x y + y 1 + x x 1 x 1 x + y 1 + y x x + y + y x x + y + y x x 1 + y + y 1 So Slope of the line P R y1 + y y y x + x x x 1 x 1 + x x x + y + y y y 1 Slope of the line QT y + y y y 1 x + x x 1 x x + x x x 1 + y + y y 1 y Slope of the line S S x 1 + x x x + y + y y y 1 y1 + y y y x + x x x 1 Slope of the line S 1 S x + x x x 1 + y + y y 1 y y + y y y 1 x + x x 1 x Now it is clear that slope of P R slope of S S 1 slope of Q T slope of S 1 S. Hence S 1 S Q T S S P R. Theorem.. Let V 1 V V and V are the points of intersection of the lines PR QT S 1 S and S S then the four points V 1 V V and V are concyclic see Fig.. Proof. From Theorem.1 it is clear that V 1 V V V and V V V V 1. Hence the four points V 1 V V and V are concyclic which completes the proof of the Theorem.. Theorem.. Let V 1 V V and V are the points of intersection of the lines P R Q T S 1 S and S S then the four points V 1 V V and V are concyclic see Fig.. Proof. From Theorem. it is clear that V 1 V V V and V V V V 1. Hence the four points V 1 V V and V are concyclic which completes the proof of the Theorem..

6 A note on special cases of Van Aubel s theorem Fig.. Fig.. Theorem.5. The quadrilaterals PQ RT P QR T S 1 S S S and S 1 S S S are parallelograms. Proof. To prove the quadrilateral PQ RT P QR TS 1 S S S S 1 S S S are parallelograms It is enough to prove that diagonals bisect each other. It is clear that The mid point of PR The mid point of Q T M 1 x1 + x + x + x + y 1 y + y y y1 + y + y + y x1 x + x x

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 7 The mid point of QT The mid point of P R x1 + x + x + x y 1 y + y y y1 + y + y + y + x1 x + x x M The mid point of S 1 S The mid point of S S y1 y + y y + x1 + x + x + x M x 1 x + x x + y 1 + y + y + y The mid point of S S The mid point of S 1 S y1 y + y y + x1 + x + x + x M x 1 x + x x + y 1 + y + y + y Hence Theorem.5 is proved. Theorem.6. Let M 1 M M M are the point of intersections of the diagonals of the parallelograms PQ RT P QR T S 1 S S S and S 1 S S S respectively then M 1 M M M are collinear and they lies on the line for recognisation sake let us call this line as van aubel s line given by x 1 x + x x x + y 1 y + y y y x 1 + x x + x + y 1 + y y + y Proof. Consider λ x 1 + x + x + x β y 1 + y + y + y γ x 1 x + x x and δ y 1 y + y y. So x1 + x + x + x + y 1 y + y y y1 + y + y + y x1 x + x x λ + δ M 1 β γ x1 + x + x + x y 1 y + y y y1 + y + y + y + x1 x + x x λ δ M β + γ y1 y + y y + x1 + x + x + x M x 1 x + x x + y 1 + y + y + y δ + λ γ + β and y1 y + y y + x1 + x + x + x M x 1 x + x x + y 1 + y + y + y δ + λ γ + β Consider a line γx + δy λγ + βδ 1 Clearly the four points M 1 M M and M lies on this line 1. Hence the four points M 1 M M and M are collinear. The line through these points is γx + δy λγ + βδ. That is x 1 x + x x x + y 1 y + y y y x 1 + x x + x + y 1 + y y + y Remark.1. 1. It is clear that the Mid Point of M 1 M the Mid Point of M M M λ β x 1 +x +x +x y 1+y +y +y.. The point M is also the point of intersection of diagonals of the parallelograms formed by joining the midpoints of the quadrilaterals PQRS and P Q R S.. For recognization sake let us call the point M as Van Aubel s point of the quadrilateral ABC D. The point M acts as midpoint of the diagonals for any arbitrary parallelogram rectangle rhombus square

8 A note on special cases of Van Aubel s theorem Fig. 5.. Using Theorem.5 and Theorem.6 it can also be stated as The midpoints of PR M 1 S 1 S M QT M S S M are collinear and M is the midpoint of M 1 M and M M. In the similar manner the midpoints of P R M S 1 S M Q T M 1 S S M are collinear and M is the midpoint of M 1 M and M M 5. Using Theorem.5 and Theorem.6 we can see how to locate the point M using only ruler and compass If some arbitrary quadrilateral ABC D is given construct the equilateral triangles on the sides either inside or outside Let PQRT be its affix vertices locate the midpoints of the sides of quadrilateral PQRT then the point of intersection of the diagonals of quadrilateral formed by the midpoints of sides of PQRT is required M. see Fig. 5 6. If I 1 I I I and O 1 O O O and G 1 G G G are incentres circumcenters and centroids of the triangles AB MBC MC DM and D AM respectively then the sets {I 1 I I I } and {O 1 O O O }and {G 1 G G G } are con cyclic when ABC D is kite or square. The orthocenters H 1 H H H of the triangles AB MBC MC DM and D AM are collinear when ABC D is kite and coincides with M when ABC D is square see Fig. 6..1. Generalizations Theorem.7. If S 1 S S S are the circumcenters of the isosceles triangles ABP BCQ C DR D AT whose base angle is θ constructed entirely out wards on the sides of quadrilateral ABCD Then a The midpoints of PR M 1 S 1 S M QT M S S M are collinear and lie on the van Aubel s line given by x 1 x + x x x + y 1 y + y y y x 1 + x x + x + y 1 + y y + y b Van Aubel s point M is the midpoint of M 1 M and M M see figure-7 Proof. We have by.1 the coordinates of PQRS are given by x1 + x + tanθ y 1 y P y1 + y tanθ x1 x

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 9 Fig. 6. x + x + tanθ y y Q x + x + tanθ y y R x + x 1 + tanθ y y 1 T y + y tanθ x x y + y tanθ x x y + y 1 tanθ x x 1 And using. the circumcenters S 1 S S and S are given by x1 + x cotθ y 1 y y1 + y + cotθ x1 x S 1 x + x cotθ y y y + y + cotθ x x S x + x cotθ y y y + y + cotθ x x S x + x 1 cotθ y y 1 y + y 1 + cotθ x x 1 S The mid point of PR x1 + x + x + x + tanθ y 1 y + y y y1 + y + y + y tanθ x1 x + x x M 1 The mid point of QT x1 + x + x + x tanθ y 1 y + y y y1 + y + y + y + tanθ x1 x + x x M The mid point of S 1 S x1 + x + x + x cotθ y 1 y + y y y1 + y + y + y + cotθ x1 + x + x + x M

0 A note on special cases of Van Aubel s theorem Fig. 7. The mid point of S S x1 + x + x + x + cotθ y 1 y + y y y1 + y + y + y cotθ x1 + x + x + x M Hence M The Midpoint of M 1 M The Mid point of M M x1 + x + x + x y 1 + y + y + y Hence b is proved. Now to prove a consider λ x 1 + x + x + x β y 1 + y + y + y γ x 1 x + x x and δ y 1 y + y y then λ + tanθ δ M 1 β tanθ γ λ tanθ δ M β + tanθ γ λ cotθ δ M β + cotθ γ and λ + cotθ δ M β cotθ γ Consider a line γx + δy λγ + βδ Clearly the four points M 1 M M and M lies on this line. Hence the four points M 1 M M and M are collinear. From Theorem.7 Clearly the line is Van Aubel s line L. Its equation is given by x 1 x + x x x + y 1 y + y y y x 1 + x x + x + y 1 + y y + y

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 1 Theorem.8. If S 1 S S S are the circumcenters of an isosceles triangles ABP BCQ C DR D AT whose base angle is θ constructed entirely inwards on the sides of quadrilateral ABCD. Then a The midpoints of P R M 1 S 1 S M Q T M S S M are collinear and lies on the Van Aubel s Line L b Van Aubel s point M is the midpoint of M 1 M and M M Proof. We have by.1 the coordinates of P Q R T are given by P x1 + x tanθ y 1 y y1 + y + tanθ x 1 x Q x + x tanθ y y y + y + tanθ x x R x + x tanθ y y T x + x 1 tanθ y y 1 y + y + tanθ x x y + y 1 + tanθ x x 1 And using. the circumcenters S 1 S S and S are given by S x1 1 + x + cotθ y 1 y y1 + y cotθ x 1 x S x + x + cotθ y y y + y cotθ x x S x + x + cotθ y y y + y cotθ x x S x + x 1 + cotθ y y 1 y + y 1 cotθ x x 1 The mid point of P R M x1 1 + x + x + x tanθ y 1 y + y y y1 + y + y + y + tanθ x 1 x + x x The mid point of Q T M x1 + x + x + x + tanθ y 1 y + y y y1 + y + y + y tanθ x 1 x + x x The mid point of S 1 S M x1 + x + x + x + cotθ y 1 y + y y y1 + y + y + y cotθ x 1 + x + x + x The mid point of S S M x1 + x + x + x cotθ y 1 y + y y y1 + y + y + y + cotθ x 1 + x + x + x Hence M The Midpoint of M 1 M The Mid point of M M x1 + x + x + x y 1 + y + y + y Hence b is proved. Now to prove a Consider λ x 1 + x + x + x β y 1 + y + y + y γ x 1 x + x x and δ y 1 y + y y. Then λ tanθ M 1 δ β + tanθ γ λ + tanθ M δ β tanθ γ

A note on special cases of Van Aubel s theorem λ + cotθ M δ β cotθ γ λ cotθ M δ β + cotθ γ Consider a line γx + δy λγ + βδ Clearly the four points M 1 M M and M lies on this line. Hence collinear. The line through these points is γx + δy λγ + βδ. From Theorem.7 Clearly the line is Van Aubel s line L. The four points M 1 M M and M are Remark.. 1. The Van Aubel s point M of the quadrilateral ABCD and the points M 1 M M M M 1 M M and M all lie on the Van Aubel s Line of the quadrilateral ABC D.. If θ and θ of Theorem.7 and Theorem.8 are equal Then the points M 1 M M M respectively coincide with the points M 1 M M and M. Theorem.9. The quadrilaterals PQ RT P QR T S 1 S S S and S 1 S S S are parallelograms where P Q R T P Q R T are the affixes of the isosceles triangles with base angles θ constructed on the sides of the quadrilateral ABCD out and inwards respectively. Proof. To prove the quadrilateral PQ RT P QR TS 1 S S S S 1 S S S are parallelograms it is enough to prove that diagonals bisect each other. By Theorem.7 and Theorem.8 it is clear that The mid point of PR The mid point of Q T x1 + x + x + x + tanθ y 1 y + y y y1 + y + y + y tanθ x1 x + x x The mid point of QT The mid point of P R x1 + x + x + x tanθ y 1 y + y y y1 + y + y + y + tanθ x1 x + x x The mid point of S 1 S The mid point of S S x1 + x + x + x cotθ y 1 y + y y y1 + y + y + y + cotθ x1 + x + x + x The mid point of S S The mid point of S 1 S x1 + x + x + x + cotθ y 1 y + y y y1 + y + y + y cotθ x1 + x + x + x Hence theorem is proved. Theorem.10. Suppose ABCD is a given arbitrary quadrilateral let P 1 P...P k+1 Q 1 Q...Q k+1 R 1 R...R k+1 T 1 T...T k+1 P 1 P...P k+1 Q 1 Q...Q k+1 R 1 R...R k+1 and T 1 T...T be the regular polygons of k + 1 sides constructed on the k+1 sides of ABCD out and inwards respectively where k 1 such that P 1 P k+1 AB P 1 P k+1 Q 1Q k+1 BC Q 1 Q k+1 R 1R k+1 C D R 1 R k+1 T 1T k+1 D A T 1 T k+1 and S 1 S S S S 1 S S S are the centers of the regular polygons constructed on the sides then a The midpoints of P k+1 Q k+1 M 1 S 1 S M R k+1 T k+1 M S S M P k+1 Q k+1 M 1 S 1 S M R k+1 T k+1 M S S M are collinear and lie on the van Aubel s linel given by x 1 x + x x x + y 1 y + y y y x 1 + x x + x + y 1 + y y + y

b The quadrilaterals P k+1 Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 Q k+1 R k+1 T k+1 P k+1 Q k+1 R k+1 T k+1 S 1 S S S and S 1 S S S are parallelograms. Proof. It is clear that in the regular polygons P 1 P...P k+1 Q 1 Q...Q k+1 R 1 R...R k+1 T 1 T...T k+1 P 1 P...P k+1 Q 1 Q...Q k+1 R 1 R...R k+1 and T 1 T...T the triangles AP K +1 BBQ K +1 C C R K +1 DDT K +1 A are isosceles triangles k+1 with base angle θ constructed outwards on the sides ABBC C DD A of quadrilateral ABC D here as the triangles AP K +1 BBQ K +1 C C R K +1 DDT K +1 A are also isosceles triangles with base angle θ constructed inwards on the sides ABBC C DD A of quadrilateral ABC D. Hence By Theorem.7 and.8 a is true. In the similar manner we can prove b using Theorem.9. Remark.. Clearly by Theorem.10 it is true that we can also plot Van Aubel s Point M for an arbitrary quadrilateral ABC D by constructing the regular polygons of n number of sides on the sides of quadrilateral lie inwards or outwards and by applying the procedure discussed in.1. Theorem.11. Let ABCD is a quadrilateral Suppose the triangles ABP C DR are isosceles with angle at their top vertices and BCQ D AT are isosceles with angle π at their top vertices all of them have same orientation constructed on the sides of the quadrilateral which lie outwards and if S 1 S S S the circumcenters of the triangles ABP BCQ C DR and D AT then a PR is perpendicular to QT. b The ratio of these two segements PR and QT doesn t depend from the quadrilateral. c Quadrilateral S 1 S S S is parallelogram. d The three points Van Aubel s Point M S1 S S S of quadrilateral S 1 S S S and the mid points of PRM PR and QT M QT are collinear and lie on the line Van Aubel s Line given by x 1 x + x x x + y 1 y + y y y x 1 + x x + x + y 1 + y y + y And in particular the midpoint of PRMP R and QT M QT is the Van Aubel s point of S 1 S S S.see Fig. 8 Proof. Given at the top vertices P R makes an angle Îś so the two isosceles triangles ABP C DR having the base angle as 90 o /. Using.1 we have x1 + x + cot y1 y y1 + y cot x1 x P R x + x + cot y y y + y cot x x And given at the top vertices QT makes an angle π so the two isosceles triangles BCQ D AT having the base angle as /. Hence using. we have x + x + tan y y y + y tan x x Q T x + x 1 + tan y y 1 y + y 1 tan x x 1 Now y1 + y y y cot x1 x x + x slope of PR x 1 + x x x + cot K v let y1 y y + y L v y + y y y 1 tan x x x + x 1 slope of QT x + x x x 1 + tan M v let y y y + y 1 N v x1 + x x x + cot y1 y y + y y1 + y y y cot L v x1 x x + x K v

A note on special cases of Van Aubel s theorem Fig. 8. Here it is clear that slope of PRslope of QT 1 M v K v + L v N v 0 That is PR QT Hence a is proved Now for b Consider K v y 1 + y y y cot / x1 x x + x L v x 1 + x x x + cot / y 1 y y + y Mv y + y y y 1 tan / x x x + x 1 N v x + x x x 1 tan / y y y + y 1 It is clear that K v cot / N v and L v cot / M v From it is clear that M v N v L v K v Lv + K v M v + N v K v L v cot/ N v M v Now PR QT Lv + K v M v + N v K v N v L v cot/ M v That is the ratio of two segments PR and QT doesn t depend on the quadrilateral. Hence b is proved. Now for c we proceed as follows: Since the base angles of isosceles triangles ABP C DR are 90 o / So using. we have x1 + x + cot y 1 y y1 + y cotx1 x S 1 and x + x + cot y y S y + y cotx x

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 5 In the similar manner since the base angles of isosceles triangles BCQ D AT are /. So using. we have x + x cot y y y + y + cotx x S and x + x 1 cot y y 1 S y + y 1 + cotx x 1 Now It is clear that The mid point of S 1 S The midpoint of S S x1 + x + x + x + cot y 1 y + y y y1 + y + y + y cotx1 x + x x Hence the quadrilateral S 1 S S S is parallelogram which completes the proof of c. Now for d Since the quadrilateral S 1 S S S is parallelogram Van Aubel s point M S1 S S S of quadrilateral S 1 S S S is the midpoint of the diagonals. Hence Van Aubel s point M S1 S S S of quadrilateral S1 S S S x1 + x + x + x + cot y 1 y + y y y1 + y + y + y cotx1 x + x x M S1 S S S and Mid point of PR x1 + x + x + x + cot y1 y + y y y1 + y + y + y cot x1 x + x x M PR Mid point of QT y1 M QT x 1 + x + x + x tan θ y + y y y1 + y + y + y + tan θ x 1 x + x x Consider λ x 1 + x + x + x β y 1 + y + y + y γ x 1 x + x x and δ y 1 y + y y then λ + cotδ M S1 S S S β cotγ λ + cot δ M PR β cot γ λ tan δ M QT β + tan γ The midpoint of M M PR and M QT λ + cot tan δ β cot tan γ 8 8 cot tan cot λ + cotδ β cotγ M S1 S S S Hence the midpoint of PR M PR and QT M QT is the Van Aubel s point S1 S S S. Now consider a line that is γx + δy λγ + βδ. Clearly the three points lies on this line. Hence the three points M S1 S S S M PR M QT are collinear. The line through these points is γx + δy λγ + βδ. From Theorem.7 Clearly the line is van aubel s line L. Theorem.1. Let ABCD is a quadrilateral suppose the triangles ABP C DR are isosceles with angle Îś at their top vertices and BCQ D AT are isosceles with angle π at their top vertices all of them have same orientation constructed on the sides of the quadrilateral which lie inwards and if S 1 S S S the circumcenters of the triangles ABP BCQ C DR and D AT Then

6 A note on special cases of Van Aubel s theorem a P R is perpendicular to Q T b The ratio of these two segements P R and Q T doesn t depend from the quadrilateral. c Quadrilateral S 1 S S S is parallelogram d The three points the Van Aubel s point M S 1 S S S and Q T M Q T are collinear and lie on the line Van Aubel s Line given by of quadrilateral S 1 S S S and mid points of P R M P R x 1 x + x x x + y 1 y + y y y x 1 + x x + x + y 1 + y y + y And in particular the midpoint of P R M P R and Q T M Q T is the Van Aubel s point of S 1 S S S Proof. Given at the top vertices P R makes an angle Îś so the two isosceles triangles ABP C DR having the base angle as 90 o /. Using.1 we have P R x1 + x cot y1 y x + x cot y y y1 + y + cot y + y + cot x1 x x x And given at the top vertices Q T makes an angle π so the two isosceles triangles BCQ D AT having the base angle as /. Hence using. we have Q T x + x tan y y x + x 1 tan y y 1 y + y + tan y + y 1 + tan x x x x 1 Now y1 Slope of P R + y y y + cot x1 x x + x x 1 + x x x cot y1 y y + y y Slope of Q T + y y y 1 + tan x x x + x 1 x + x x x 1 tan y y y + y 1 x1 + x x x cot y1 y y + y y1 + y y y + cot x1 x x + x L v K v K v L v let M v l et N v It is clear that slope of P R slope of Q T 1 M v K v + L v N v 0 5 That is P R Q T Hence a is proved. Now for b Consider K v y 1 + y y y + cot/x1 x x + x L v x 1 + x x x cot/ y 1 y y + y M v y + y y y 1 + tan/x x x + x 1 N v x + x x x 1 + tan/ y y y + y 1 It is clear that K v cot/ N v K v cot/ N v and L v cot/ M v From 5 it is clear that M v L v N v K Lv + K v v M K v v + N v N L v cot/ v M v

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 7 Now P R Q T L + K M + N K N L cot/ M That is the ratio of two segments P R and Q T doesn t depend from the quadrilateral. Hence b is proved. Now for c we proceed as follows: Since the base angles of isosceles triangles ABP C DR are 90 o / So using. we have S x1 1 + x cot y 1 y y1 + y + cotx1 x and S x + x cot y y y + y + cotx x In the similar manner since the base angles of isosceles triangles BCQ D AT are /. So using. we have S x + x + cot y y y + y cotx x and S x + x 1 + cot y y 1 y + y 1 cotx x 1 Now It is clear that The mid point of S 1 S The midpoint of S S x1 + x + x + x cot y 1 y + y y y1 + y + y + y + cotx1 x + x x Hence the quadrilateral S 1 S S S is parallelogram which completes the proof of c. Now for d Since the quadrilateral S 1 S S S is parallelogram Van Aubel s point S 1 S S S is the midpoint of the diagonals. Hence Van Aubel s point M x1 + x + x + x cot y 1 y + y y S 1 S S S M S 1 S S S M S 1 S S S of quadrilateral S 1 S S S y1 + y + y + y + cotx1 x + x x And midpoint of P R x1 M + x + x + x cot y1 y + y y y1 + y + y + y + cot x1 x + x x P R Mid point of Q T x1 M + x + x + x + tan y1 y + y y y1 + y + y + y tan x1 x + x x Q T Consider λ x 1 + x + x + x β y 1 + y + y + y γ x 1 x + x x δ y 1 y + y y then λ cotδ M S β + cotγ 1 S S S λ cot M δ P R β + cot γ λ + tan M δ Q T β tan γ of quadrilateral The midpoint of M P R and M Q T λ cot tan δ β + cot tan γ 8 8 λ cotδ β + cotγ M S cot tan cot 1 S S S Hence the midpoint of P R M P R and Q T M Q T is the Van Aubel s point of S 1 S S S Now consider a line that is γx + δy λγ + βδ. Clearly the three points lies on this line. Hence the three points M S 1 S M S S P R M Q T are collinear. The line through these points is γx + δy λγ + βδ. From Theorem.7 Clearly the line is van aubel s line L.

8 A note on special cases of Van Aubel s theorem Remark.. 1. The Van Aubel s point M S1 S S S of quadrilateral S1 S S S the Van Aubel s point M S of 1 S S S quadrilateral S 1 S S S and the four points M PR M QT M P R M Q T are collinear lie on the Van Aubel s Line.. Using.. it is clear that Van Aubel s line contains 15 points They are Van Aubel s point M of the quadrilateral ABCD the points M 1 M M M M 1 M M M the Van Abel s point of quadrilateral S 1S S S the Van Abel s point of quadrilateral S 1 S S S M PR M QT M P R M Q T. Theorem.1. The following statements are true. a The Van Aubel s point M of quadrilateral ABCD is the midpoint of Van Aubel s points of the quadrilaterals S 1 S S S and S 1 S S S M PR M P R and M QT M Q T see Fig. 9 b M PR M Q T M S1 S S S M S 1 S S S M QT M P R Fig. 9. Proof. We have λ + cotδ M S1 S S S β cotγ λ + cot δ M PR β cot γ λ tan δ M QT β + tan γ λ cotδ M S β + cotγ 1 S S S λ cot M δ P R β + cot γ

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 9 λ + tan M δ Q T β tan γ Now it is clear that the Van Aubel s point M of quadrilateral ABCD is the midpoint of the Van Aubel s points of the quadrilaterals S 1 S S S and S 1 S S S M PR M P R M QT M. Q T Hence a is proved. Now for b Consider M PR M cot tan Q T δ + γ cot δ + γ Hence M S1 S S S M S cot 1 S S S δ + γ M QT M cot tan P R δ + γ cot δ + γ M PR M Q T M S1 S S S M S 1 S S S M QT M P R.. Dao s Generalization [1] Theorem.1. Let ABCD be a quadrilateral let four points A 1 B 1 C 1 D 1 on the plane either interior or exterior to the quadrilateral such that A 1 AB D AD 1 B 1 BC AB A 1 β BC B 1 C 1 C D γ C DC 1 D 1 D A δ and + γ β + δ π in the same a A 1 B 1 C 1 D 1 is an orthodiagonal quadrilateral. b The ratio of these two segments A 1 C 1 and B 1 D 1 doesn t depend from the quadrilateral. Fig. 10. Proof. Without loss of generality let us consider the coordinates of vertices of the quadrilateral ABC D as A x1 y 1 B x y C x y and D x y and also given A1 AB D AD 1 B 1 BC AB A 1 β BC B 1 C 1 C D γ 90 0 C DC 1 D 1 D A δ 90 0 β. So using.1 we have the coordinates of A 1 B 1 C 1 D 1 as follows x1 cotβ + x cot ± y 1 y y1 cotβ + y cot x 1 x A 1 cot + cotβ cot + cotβ x cotγ + x cotβ ± y y y cotγ + y cotβ x x B 1 cotβ + cotγ cotβ + cotγ x + x cotcotβ ± cot y y y + y cotcotβ cotx x since γ 90 0 1 + cotcotβ 1 + cotcotβ

50 A note on special cases of Van Aubel s theorem x cotδ + x cotγ ± y y y cotδ + y cotγ x x C 1 cotγ + cotδ cotγ + cotδ x cot + x cotβ ± cotcotβ y y y cot + y cotβ cotcotβx x since γ 90 0 δ 90 0 β cot + cotβ cot + cotβ x cot + x 1 cotδ ± y y 1 y cot + y 1 cotδ x x 1 D 1 cotδ + cot cotδ + cot x cotcotβ + x 1 ± cotβ y y 1 y cotcotβ + y 1 cotβx x 1 since δ 90 0 β 1 + cotcotβ 1 + cotcotβ Now Slope of A 1 C 1 coty y + cotβy y 1 [ cotcotβx x x 1 x ] cotx x + cotβx x 1 ± [ cotcotβy y y 1 y ] K d L d Slope of B 1 D 1 y 1 y + y y cotcotβ [ cotβx x 1 cotx x ] x 1 x + x x cotcotβ ± [ cotβy y 1 coty y ] M d N d cotx x + cotβx x 1 ± [ cotcotβy y y 1 y ] coty y + cotβy y 1 [ cotcotβx x x 1 x ] L d K d So it is clear that Slope of A 1 C 1 Slope of B 1 D 1 1 M d K d + L d N d 0 6 Hence A 1 C 1 B 1 D 1 that is quadrilateral A 1 C 1 B 1 D 1 is orthodiagonal quadrilateral. So a is proved. Now for b Consider K d coty y + cotβy y 1 [ cotcotβx x x 1 x ] L d cotx x + cotβx x 1 ± [ cotcotβy y y 1 y ] M d y 1 y + y y cotcotβ [ cotβx x 1 cotx x ] N d x 1 x + x x cotcotβ ± [ cotβy y 1 coty y ] It is clear that K d N d and L d M d. From 6 it is clear that M d N d L d K d Ld + K d M d + N d K d N d L d M d 1 Now A 1 C 1 B 1 D cotcotβ + 1 Ld + K d 1 cot + cotβ M cotcotβ + 1 d + N d cot + cotβ Kd N d cotcotβ + 1 cot + cotβ Ld M d cotcotβ + 1 cot + cotβ That is the ratio of two segments PR and QT doesn t depend from the quadrilateral. Hence b is proved... Generalization of Kiepert Hyperbola Suppose the diagonals of quadrilaterals ABC D are equal suppose the triangles ABP C DR are isosceles with angle ψ at their top vertices and the triangles BCQ D AT are isosceles with angle π ψ at their top vertices all of them have the same orientation. Then intersection of PR and QT moves along an equilateral hyperbola passing through the midpoints of diagonals and asymptotes midlines of the quadrilateral. For further generalizations refer []. Acknowledgements The author is would like to thank an anonymous referee for his/her kind comments and suggestions which lead to a better presentation of this paper.

Dasari Naga Vijay Krishna / Int. J. Adv. Appl. Math. and Mech. 5 018 0 51 51 References [1] Dao Thanh Oai Generalizations of some famous classical Euclidean geometry theorems International Journal of Computer Discovered Mathematics IJCDM 1 016 1 0. [] Dao Thanh Oai Equilateral Triangles and Kiepert Perspectors in Complex Numbers Forum Geometricorum 15 015 105 11. [] Dasari Naga Vijay Krishna A New Consequence of Van Aubels Theorem Preprints 016 016110009 doi: 10.09/preprints01611.0009.v1. [] Michael de Villiers Generalizing Van Aubel Using Duality Mathematics Magazine 7 000 0 07. [5] Michael de Villiers Dual generalizations of Van Aubel s theorem The Mathematical Gazette November 1998 05 1. [6] Yutaka Nishiyama The Beautiful Geometric Theorem Of Van Aubel International Journal of Pure and Applied Mathematics 66 1 011 71 80. [7] http://www.cut-the-knot.org/curriculum/geometry/parafromtri.shtml. [8] http://www.cut-the-knot.org/curriculum/geometry/equitrionpara.shtml. [9] http://dynamicmathematicslearning.com/vanaubel-application.html. [10] http://dynamicmathematicslearning.com/aubelparm.html. Submit your manuscript to IJAAMM and benefit from: Rigorous peer review Immediate publication on acceptance Open access: Articles freely available online High visibility within the field Retaining the copyright to your article Submit your next manuscript at editor.ijaamm@gmail.com