Difference between revisions of "Gossard perspector"

Latest revision as of 05:44, 19 January 2023

1 Gossard perspector X(402) and Gossard triangle
2 Gossard perspector of right triangle
3 Gossard perspector and Gossard triangle for isosceles triangle
4 Euler line of the triangle formed by the Euler line and the sides of a given triangle
5 Gossard triangle for triangle with angle 60
6 Gossard triangle for triangle with angle 120
7 Gossard perspector
8 Zeeman’s Generalisation
9 Paul Yiu's Generalization
10 Dao Thanh Oai's Generalization

Gossard perspector X(402) and Gossard triangle

In $1765$ Leonhard Euler proved that in any triangle, the orthocenter, circumcenter and centroid are collinear. We name this line the Euler line. Soon he proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is in parallel with the third side of the given triangle.

Professor Harry Clinton Gossard in $1916$ proved that three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle which is perspective with the given triangle and has the same Euler line. The center of the perspective now is known as the Gossard perspector or the Kimberling point $X(402).$ It is the crosspoint of Gauss line and Euler line.

Let triangle $\triangle ABC$ be given. The Euler line crosses lines $AB, BC,$ and $AC$ at points $D, E,$ and $F.$

On $13/01/2023$ it was found that the Gossard perspector is the centroid of the points $A, B, C, D, E, F.$

vladimir.shelomovskii@gmail.com, vvsss

Gossard perspector of right triangle

It is clear that the Euler line of right triangle $ABC (\angle A = 90 ^\circ)$ meet the sidelines $BC, CA$ and $AB$ of $\triangle ABC$ at $A'$ and $A,$ where $A'$ is the midpoint of $BC.$

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle AA'B, \triangle AA'C,$ and the line $l$ contains $A$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle AA'C$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle AA'B$ and $l.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of $\triangle ABC.$

Let $\triangle ABC$ be any right triangle and let $\triangle A'B'C'$ be its Gossard triangle. Then the lines $AA', BB',$ and $CC'$ are concurrent. We call the point of concurrence $Go$ as the Gossard perspector of $\triangle ABC.$

$Go$ is the midpoint of $AA'.$ $A$ is orthocenter of $\triangle ABC, A'$ is circumcenter of $\triangle ABC,$ so $Go$ is midpoint of $OH.$

$M = A'C' \cap AB$ is the midpoint $AB, N = A'B' \cap AC$ is the midpoint $AC, \triangle A'BM = \triangle CNA' \sim \triangle CBA$ with coefficient $k = \frac {1}{2}.$

Any right triangle and its Gossard triangle are congruent.

Any right triangle and its Gossard triangle have the same Euler line.

The Gossard triangle of the right $\triangle ABC$ is the reflection of $\triangle ABC$ in the Gossard perspector.

vladimir.shelomovskii@gmail.com, vvsss

Gossard perspector and Gossard triangle for isosceles triangle

It is clear that the Euler line of isosceles $\triangle ABC (AB = AC)$ meet the sidelines $BC, CA$ and $AB$ of $\triangle ABC$ at $A'$ and $A,$ where $A'$ is the midpoint of $BC.$

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle AA'B, \triangle AA'C,$ and the line $l$ contains $A$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle AA'C$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle AA'B$ and $l.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of $\triangle ABC.$

Let $\triangle ABC$ be any isosceles triangle and let $\triangle A'B'C'$ be its Gossard triangle. Then the lines $AA', BB',$ and $CC'$ are concurrent. We call the point of concurrence $Go$ as the Gossard perspector of $\triangle ABC.$ Let $H$ be the orthocenter of $\triangle ABC, O$ be the circumcenter of $\triangle ABC.$

It is clear that $Go$ is the midpoint of $AA'.$ $M = A'C' \cap AB$ is the midpoint $AB, N = A'B' \cap AC$ is the midpoint $AC.$

$\triangle A'BM = \triangle CNA' \sim \triangle CBA$ with coefficient $k = \frac {1}{2}.$

Any isosceles triangle and its Gossard triangle are congruent.

Any isosceles triangle and its Gossard triangle have the same Euler line.

The Gossard triangle of the isosceles $\triangle ABC$ is the reflection of $\triangle ABC$ in the Gossard perspector. Denote $\angle BAC = \alpha, \angle ABC = \angle ACB = \beta, AO = BO = R \implies$ $\sin \beta = \cos \frac {\alpha}{2}, GoO = AO – AGo = R \cdot \frac{1 – \cos {\alpha}}{2},$ $OH = AH – AO = R(2 \cos \alpha – 1) \implies \vec {GoO} = \vec {OH} \cdot \frac {1 – \cos \alpha}{2(2 \cos \alpha – 1)}.$

vladimir.shelomovskii@gmail.com, vvsss

Euler line of the triangle formed by the Euler line and the sides of a given triangle

Let the Euler line of $\triangle ABC$ meet the lines $AB, AC,$ and $BC$ at $D, E,$ and $F,$ respectively.

Euler line of the $\triangle ADE$ is parallel to $BC.$ Similarly, Euler line of the $\triangle BDF$ is parallel to $AC,$ Euler line of the $\triangle CEF$ is parallel to $AB.$

Proof

Denote $\angle A = \alpha, \angle B = \beta, \angle C = \gamma,$ smaller angles between the Euler line and lines $BC, AC,$ and $AB$ as $\theta_A, \theta_B,$ and $\theta_C,$ respectively. WLOG, $AC > BC > AB.$ It is known that $\tan \theta_A = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}, \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}, \tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.$

Euler line

Let $O'$ be circumcenter of $\triangle ADE, KO'$ be Euler line of $\triangle ADE, K \in DE$ (line).

Similarly, $\tan \angle O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.$ $3(\tan\alpha – \tan \gamma) (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \gamma) (3 – \tan \alpha \cdot \tan \beta) = (\tan^2 \alpha – 3) \cdot (3 –\tan \beta \cdot \tan \gamma),$ $(3 – \tan \alpha \cdot \tan \gamma) \cdot (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \beta) \cdot (\tan\alpha – \tan \gamma) = (\tan^2 \alpha – 3) \cdot (\tan \beta – \tan \gamma).$ Suppose, $\tan^2 \alpha \ne 3$ which means $\alpha \ne 60^\circ$ and $\alpha \ne 120^\circ.$ In this case $\tan \angle O'KF = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} = \tan \theta_A \implies \angle O'KF = \theta_A \implies O'K||BC.$

Similarly one can prove the claim in the other cases.

vladimir.shelomovskii@gmail.com, vvsss

Gossard triangle for triangle with angle 60

Let $\angle A$ of the triangle $ABC$ be $60^\circ, \angle B \ne 60^\circ.$ Let the Euler line of $\triangle ABC$ meet the lines $AB, AC$ and $BC$ at points $D, E,$ and $F,$ respectively. Prove that $\triangle ADE$ is an equilateral triangle.

Proof

Denote $\angle ABC = \beta, \angle ACB = \gamma.$ It is known that $\tan \angle AED = \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}.$

Euler line

$\tan \angle AED = \frac{3 – \sqrt{3} \tan \gamma}{\sqrt{3} – \tan \gamma} = \sqrt{3} \implies \angle AED = 60^\circ.$ Therefore $\triangle ADE$ is equilateral triangle.

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle BDF, \triangle CEF,$ and the line $l$ contains centroid $G$ of the $\triangle ADE$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle CEF$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle BDF$ and $l,$ the vertex $A'$ being the intersection of the Euler lines of the $\triangle BDF$ and $\triangle CEF.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of the $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Gossard triangle for triangle with angle 120

Let $\angle A$ of the triangle $ABC$ be $120^\circ, \angle B \ne 30^\circ.$ Let the Euler line of $\triangle ABC$ meet the lines $AB, AC$ and $BC$ at points $D, E,$ and $F,$ respectively. Then $\triangle ADE$ is an equilateral triangle.

One can prove this claim using the same formulae as in the case $\angle A = 60^\circ.$

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle BDF, \triangle CEF,$ and the line $l$ contains centroid $G$ of the $\triangle ADE$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle CEF$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle BDF$ and $l,$ the vertex $A'$ being the intersection of the Euler lines of the $\triangle BDF$ and $\triangle CEF.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of the $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Gossard perspector

Let non equilateral triangle $ABC$ be given. The Euler line of $\triangle ABC$ crosses lines $AB, BC,$ and $AC$ at points $D, E,$ and $F,$ respectively.

Let the point $X$ be the centroid of the set of points $A, B, C, D, E, F.$

Let Gossard triangle $A'B'C'$ be defined as described above.

Prove that $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center is the point $X,$ the Euler line of $\triangle A'B'C'$ coincide with the Euler line of $\triangle ABC.$

Proof

Denote $G_A, G_B,$ and $G_C$ centroids of the triangles $ADE, BDF,$ and $CEF,$ respectively. It is clear that $G_A \in B'C', G_B \in A'C', G_C \in A'B', X \in$ Euler line.

Let point $G'_A$ be symmetric to the point $G_A$ with respect to the point $X.$

Similarly we define points $G'_B$ and $G'_C.$ $\vec {G'_A} = 2 \vec X – \vec G_A = \frac {\vec A+\vec B+\vec C+ \vec D+\vec E + \vec F}{3} – \frac {\vec A+ \vec D+\vec E }{3} = \frac {\vec B+\vec C+ \vec F}{3} \in BC.$ Similarly $G'_B \in AC$ and $G'_C \in AB.$

$AB||A'B', AC || A'C', BC|| B'C' \implies$ the crosspoints of lines $A'B', A'C',$ and $B'C'$ are symmetric to the crosspoints of lines $AB, AC,$ and $BC,$ therefore points $A', B',$ and $C',$ are symmetric to points $A, B,$ and $C$ with respect to the point $X \implies X$ is the Gossard perspector of the $\triangle ABC.$

It is clear that the Gossard perspector lyes on Euler line of the $\triangle ABC$ and $\triangle A'B'C'$ is congruent to $\triangle ABC$ .

The Euler line of $\triangle A'B'C'$ is symmetric to the Euler line of $\triangle ABC$ with respect to $X.$ Therefore these lines coincide.

vladimir.shelomovskii@gmail.com, vvsss

Zeeman’s Generalisation

Let $l$ be any line parallel to the Euler line of non equilateral triangle $ABC.$ Let $l$ intersect the sidelines $AB, CA, BC$ of $\triangle ABC$ at points $D, E, F,$ respectively. Let $\triangle A'B'C'$ be the triangle formed by the Euler lines (as in previous sections) of the triangles $\triangle ADE, \triangle BDF,$ and $\triangle CEF.$ Let the point $X$ be the centroid of the set of points $A, B, C, D, E, F.$

Then $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center is the point $X,$ the Euler line of $\triangle A'B'C'$ coincide with the line $DE$ and the point $X$ is equidistant from the Euler lines.

In this case $X$ usually called the Zeeman–Gossard perspector.

One can prove this claim using the method of previous section. vladimir.shelomovskii@gmail.com, vvsss

Paul Yiu's Generalization

Let $P$ be any point in the plane of non equilateral triangle $ABC$ different from its centroid $G.$

Let the line $PG$ meet the sidelines $AB, CA,$ and $BC$ at $D, E,$ and $D,$ respectively.

Let the centroids of the triangles $AED, BDF,$ and $CFE$ be $Ga, Gb,$ and $Gc,$ respectively.

Let $Pa$ be a point such that $EPa$ is parallel to $CP$ and $DPa$ is parallel to $BP.$ Symilarly, $Pb: DPb||AP, FPb||CP, Pc: FPc||BP, EPc||AP.$

Let $\triangle A'B'C'$ be the triangle formed by the lines $GaPa, GbPb,$ and $GcPc.$ Let the point $X$ be the centroid of the set of points $A, B, C, D, E, F.$ Then $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center is the point $X,$ the points $P, G,$ and $G'$ are collinear.

One can prove this claim using the method of previous section.

The points $P, Pa, Pb,$ and $Pc$ are collinear.

vladimir.shelomovskii@gmail.com, vvsss

Dao Thanh Oai's Generalization

Let triangle $ABC$ and line $\ell,$ non parallel to sidelines be given. Let line $\ell$ meets sidelines $AB, AC, BC$ of $\triangle ABC$ at points $D, E, F,$ respectively.

Let $Q \in \ell.$ Let $K, L, M$ be the points such that $QA||EM||DL, QB||KD||FM, QC||KE||FL.$

Similarly define points $P \in \ell, K_0, L_0, M_0.$

Let triangle $\triangle A'B'C'$ be the triangle formed by the lines $KK_0, LL_0, MM_0.$

Prove that $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center lies on $\ell.$

Proof

Let $G$ and $G_0$ be the midpoints $BE$ and $CD,$ respectively.

Let $X$ be the crosspoint of $\ell$ and Gauss line $GG_0.$

Let $A', B', C', D', E',$ and $K'$ be the points simmetric to $A, B, C, D, E,$ and $K$ with respect to $X,$ respectively.

We will prove that $K' \in BC$ which means that $K \in B'C' \implies K_0 \in BC, L \in AC$ and so on.

$G$ midpoint $BE, X$ midpoint $EE' \implies GX||BE'.$

$G_0$ midpoint $DC, X$ midpoint $DD' \implies G_0X||CD' \implies BE'||CD'.$ $KD||K'D'||BQ, KE||K'E'||CQ, CD'||BE', Q \in D'E' \implies K' \in BC$ according the Claim.

Claim (Parallel lines in trapezium)

Let $ABCD$ be the quadrungle such that $AD||BC.$ Let $M \in AB, MD||BN, AN||CM.$ Prove that point $N$ lyes on $CD.$

Proof

We prove Claim in the case $AB$ non parallel to $CD.$ Denote $Q = AB \cap CD.$

$AD||BC \implies \frac {QB}{QA}= \frac{QC}{QD}.$

Let $BN||MD$ cross $CD$ at $N.$ Then $\frac {QN}{QD}=\frac {QB}{QM}.$

$\frac {QC}{QN} = \frac {QC}{QD} \cdot \frac {QD}{QN} = \frac {QB}{QA} \cdot \frac {QM}{QB} = \frac {QM}{QA} \implies AN||CM.$ The Claim is correct in the case of non convex $ABCD.$ One can simplify the proof of Dao Generalization using this variant of the Claim.

vladimir.shelomovskii@gmail.com, vvsss

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Difference between revisions of "Gossard perspector"

Latest revision as of 05:44, 19 January 2023

Contents

Gossard perspector X(402) and Gossard triangle

Gossard perspector of right triangle

Gossard perspector and Gossard triangle for isosceles triangle

Euler line of the triangle formed by the Euler line and the sides of a given triangle

Gossard triangle for triangle with angle 60

Gossard triangle for triangle with angle 120

Gossard perspector

Zeeman’s Generalisation

Paul Yiu's Generalization

Dao Thanh Oai's Generalization

@@ Line 1: / Line 1: @@
 ==Gossard perspector X(402) and Gossard triangle==
-Euler proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is parallel with the third side of the given triangle.
+In <math>1765</math> Leonhard Euler proved  that in any triangle, the orthocenter, circumcenter and centroid are collinear. We name this line the Euler line. Soon he proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is in parallel with the third side of the given triangle.
-Gossard proved that the three Euler lines of the triangles formed by the Euler line and the sides, taken by twos, of a given triangle, form a triangle triply perspective with the given triangle and having the same Euler line. The orthocenters, circumcenters and centroids of these two triangles are symmetrically placed as to the center of perspective which known as Gossard perspector or Kimberling point <math>X(402).</math>
+Professor Harry Clinton Gossard in <math>1916</math> proved that three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle which is perspective with the given triangle and has the same Euler line. The center of the perspective now is known as the Gossard perspector or the Kimberling point <math>X(402).</math> It is the crosspoint of Gauss line and Euler line.
+Let triangle <math>\triangle ABC</math> be given. The Euler line crosses lines <math>AB, BC,</math> and <math>AC</math> at points <math>D, E,</math> and <math>F.</math>
+On <math>13/01/2023</math> it was found that the Gossard perspector is the centroid of the points   <math>A, B, C, D, E, F.</math>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
 ==Gossard perspector of right triangle==
@@ Line 26: / Line 32: @@
 '''vladimir.shelomovskii@gmail.com, vvsss'''
-==Gossard perspector and Gossard triangle for equilateral triangle==
+==Gossard perspector and Gossard triangle for isosceles triangle==
 [[File:Gossard equilateral.png|500px|right]]
-It is clear that the Euler line of equilateral <math>\triangle ABC (AB = AC)</math> meet the sidelines <math>BC, CA</math> and <math>AB</math> of <math>\triangle ABC</math> at <math>A'</math> and <math>A,</math> where <math>A'</math> is the midpoint of <math>BC.</math>
+It is clear that the Euler line of isosceles <math>\triangle ABC (AB = AC)</math> meet the sidelines <math>BC, CA</math> and <math>AB</math> of <math>\triangle ABC</math> at <math>A'</math> and <math>A,</math> where <math>A'</math> is the midpoint of <math>BC.</math>
 Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle AA'B, \triangle AA'C,</math> and the line <math>l</math> contains <math>A</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle AA'C</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle AA'B</math> and <math>l.</math>
@@ Line 34: / Line 40: @@
 We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of <math>\triangle ABC.</math>
-Let <math>\triangle ABC</math> be any equilateral triangle and let  <math>\triangle A'B'C'</math> be its Gossard triangle. Then the lines <math>AA', BB',</math> and <math>CC'</math> are concurrent. We call the point of concurrence <math>Go</math> as  the Gossard perspector of <math>\triangle ABC.</math>  Let <math>H</math> be the orthocenter of <math>\triangle ABC, O</math> be the circumcenter of <math>\triangle ABC.</math>
+Let <math>\triangle ABC</math> be any isosceles triangle and let  <math>\triangle A'B'C'</math> be its Gossard triangle. Then the lines <math>AA', BB',</math> and <math>CC'</math> are concurrent. We call the point of concurrence <math>Go</math> as  the Gossard perspector of <math>\triangle ABC.</math>  Let <math>H</math> be the orthocenter of <math>\triangle ABC, O</math> be the circumcenter of <math>\triangle ABC.</math>
 It is clear that <math>Go</math> is the midpoint of <math>AA'.</math>
@@ Line 41: / Line 47: @@
 <math>\triangle A'BM = \triangle CNA' \sim \triangle CBA</math> with coefficient <math>k = \frac {1}{2}.</math>
-Any equilateral  triangle and its Gossard triangle are congruent.
+Any isosceles  triangle and its Gossard triangle are congruent.
-Any equilateral  triangle and its Gossard triangle have the same Euler line.
+Any isosceles  triangle and its Gossard triangle have the same Euler line.
-The Gossard triangle of the equilateral  <math>\triangle ABC</math> is the reflection of <math>\triangle ABC</math> in the Gossard perspector.
+The Gossard triangle of the isosceles  <math>\triangle ABC</math> is the reflection of <math>\triangle ABC</math> in the Gossard perspector.
 Denote <math>\angle BAC = \alpha, \angle ABC = \angle ACB = \beta, AO = BO = R \implies</math>
 <cmath>\sin \beta = \cos \frac {\alpha}{2}, GoO = AO – AGo = R \cdot \frac{1 – \cos  {\alpha}}{2},</cmath>
 <cmath>OH = AH – AO = R(2 \cos \alpha – 1)  \implies \vec {GoO} = \vec {OH} \cdot \frac {1 – \cos \alpha}{2(2 \cos \alpha – 1)}.</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Euler line of the triangle formed by the Euler line and the sides of a given triangle==
+[[File:Euler Euler line.png|500px|right]]
+Let the Euler line of <math>\triangle ABC</math> meet the lines <math>AB, AC,</math> and <math>BC</math> at <math>D, E,</math> and <math>F,</math> respectively.
+Euler line of the <math>\triangle ADE</math> is parallel to <math>BC.</math> Similarly, Euler line of the <math>\triangle BDF</math> is parallel to <math>AC,</math> Euler line of the <math>\triangle CEF</math> is parallel to <math>AB.</math>
+<i><b>Proof</b></i>
+Denote <math>\angle A = \alpha, \angle B = \beta, \angle C = \gamma,</math> smaller angles between the Euler line and lines <math>BC, AC,</math> and <math>AB</math> as <math>\theta_A, \theta_B,</math> and <math>\theta_C,</math> respectively. WLOG, <math>AC > BC > AB.</math>
+It is known that <math>\tan \theta_A =  \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}, \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}, \tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.</math>
+*[[Euler line]]
+Let <math>O'</math> be circumcenter of <math>\triangle ADE, KO'</math> be Euler line of <math>\triangle ADE, K \in DE</math> (line).
+Similarly, <math>\tan \angle O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.</math>
+<cmath>3(\tan\alpha – \tan \gamma) (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \gamma) (3 – \tan \alpha \cdot \tan \beta) = (\tan^2 \alpha – 3) \cdot (3 –\tan \beta \cdot \tan \gamma),</cmath>
+<cmath>(3 – \tan \alpha \cdot \tan \gamma) \cdot (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \beta) \cdot (\tan\alpha – \tan \gamma) =  (\tan^2 \alpha – 3) \cdot (\tan \beta – \tan \gamma).</cmath>
+Suppose, <math>\tan^2 \alpha \ne  3</math> which means <math>\alpha \ne 60^\circ</math> and <math>\alpha \ne 120^\circ.</math> In this case
+<cmath>\tan \angle O'KF =  \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} = \tan \theta_A \implies \angle O'KF = \theta_A \implies O'K||BC.</cmath>
+Similarly one can prove the claim in the other cases.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Gossard triangle for triangle with angle 60==
+[[File:Gossard 60.png|500px|right]]
+Let <math>\angle A</math> of the triangle <math>ABC</math> be <math>60^\circ, \angle B \ne 60^\circ.</math>
+Let the Euler line of <math>\triangle ABC</math> meet the lines <math>AB, AC</math> and <math>BC</math> at points <math>D, E,</math> and <math>F,</math> respectively.
+Prove that <math>\triangle ADE</math> is an equilateral triangle.
+<i><b>Proof</b></i>
+Denote <math>\angle ABC = \beta, \angle ACB = \gamma.</math>
+It is known that <cmath>\tan \angle AED = \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}.</cmath>
+*[[Euler line]]
+<cmath>\tan \angle AED = \frac{3 – \sqrt{3} \tan \gamma}{\sqrt{3} – \tan \gamma} = \sqrt{3} \implies \angle AED = 60^\circ.</cmath>
+Therefore <math>\triangle ADE</math> is equilateral triangle.
+Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle BDF, \triangle CEF,</math> and the line <math>l</math> contains centroid <math>G</math> of the <math>\triangle ADE</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle CEF</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle BDF</math> and <math>l,</math> the vertex <math>A'</math> being the intersection of the Euler lines of the <math>\triangle BDF</math> and <math>\triangle CEF.</math>
+We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of the <math>\triangle ABC.</math>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Gossard triangle for triangle with angle 120==
+[[File:Gossard 120.png|500px|right]]
+Let <math>\angle A</math> of the triangle <math>ABC</math> be <math>120^\circ, \angle B \ne 30^\circ.</math>
+Let the Euler line of <math>\triangle ABC</math> meet the lines <math>AB, AC</math> and <math>BC</math> at points <math>D, E,</math> and <math>F,</math> respectively. Then  <math>\triangle ADE</math> is an equilateral triangle.
+One can prove this claim using the same formulae as in the case <math>\angle A = 60^\circ.</math>
+Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle BDF, \triangle CEF,</math> and the line <math>l</math> contains centroid <math>G</math> of the <math>\triangle ADE</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle CEF</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle BDF</math> and <math>l,</math> the vertex <math>A'</math> being the intersection of the Euler lines of the <math>\triangle BDF</math> and <math>\triangle CEF.</math>
+We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of the <math>\triangle ABC.</math>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Gossard perspector==
+[[File:Gossard complete.png|450px|right]]
+Let non equilateral triangle <math>ABC</math> be given. The Euler line of <math>\triangle ABC</math> crosses lines <math>AB, BC,</math> and <math>AC</math> at points <math>D, E,</math> and <math>F,</math> respectively.
+Let the point <math>X</math> be the centroid of the set of points <math>A, B, C, D, E, F.</math>
+Let Gossard triangle <math>A'B'C'</math> be defined as described above.
+Prove that <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center is the point <math>X,</math> the Euler line of <math>\triangle A'B'C' </math> coincide with the Euler line of <math>\triangle ABC.</math>
+<i><b>Proof</b></i>
+Denote <math>G_A, G_B,</math> and <math>G_C</math> centroids of the triangles <math>ADE, BDF,</math> and <math>CEF,</math> respectively.
+It is clear that <math>G_A \in B'C', G_B \in A'C', G_C \in A'B', X \in</math> Euler line.
+Let point <math>G'_A</math> be symmetric to the point <math>G_A</math> with respect to the point <math>X.</math>
+Similarly we define points <math>G'_B</math> and <math>G'_C.</math>
+<cmath>\vec {G'_A} = 2 \vec X – \vec G_A = \frac {\vec A+\vec B+\vec C+ \vec D+\vec E + \vec F}{3} – \frac {\vec A+ \vec D+\vec E }{3} =  \frac {\vec B+\vec C+ \vec F}{3} \in BC.</cmath>
+Similarly <math>G'_B \in AC</math> and <math>G'_C \in AB.</math>
+<math>AB||A'B', AC || A'C', BC|| B'C' \implies</math> the crosspoints of lines <math>A'B', A'C',</math> and <math>B'C'</math> are symmetric to the crosspoints of lines <math>AB, AC,</math> and <math>BC,</math> therefore points <math>A', B',</math> and <math>C',</math> are symmetric to points <math>A, B,</math> and <math>C</math> with respect to the point <math>X \implies X</math> is the Gossard perspector of the <math>\triangle ABC.</math>
+It is clear that the Gossard perspector lyes on Euler line of the <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> is congruent to <math>\triangle ABC</math>.
+The Euler line of <math>\triangle A'B'C'</math> is symmetric to the Euler line of <math>\triangle ABC</math> with respect to <math>X.</math> Therefore these lines coincide.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Zeeman’s Generalisation==
+[[File:Generalization 1.png|500px|right]]
+Let <math>l</math> be any line parallel to the Euler line of non equilateral triangle <math>ABC.</math> Let <math>l</math> intersect the sidelines <math>AB, CA, BC</math> of <math>\triangle ABC</math> at points <math>D, E, F,</math> respectively. Let <math>\triangle A'B'C'</math>  be the triangle formed by the Euler lines (as in previous sections) of the triangles <math>\triangle ADE, \triangle BDF,</math> and <math>\triangle CEF.</math> Let the point <math>X</math> be the centroid of the set of points <math>A, B, C, D, E, F.</math>
+Then <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center is the point <math>X,</math> the Euler line of <math>\triangle A'B'C'</math> coincide with the  line <math>DE</math> and the point <math>X</math> is equidistant from the Euler lines.
+In this case <math>X</math> usually called the Zeeman–Gossard perspector.
+One can prove this claim using the method of previous section.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Paul Yiu's Generalization==
+[[File:Yu generalization.png|500px|right]]
+Let <math>P</math> be any point in the plane of non equilateral triangle <math>ABC</math> different from its centroid <math>G.</math>
+Let the line <math>PG</math> meet the sidelines <math>AB, CA,</math> and <math>BC</math> at <math>D, E,</math> and <math>D,</math> respectively.
+Let the centroids of the triangles <math>AED, BDF,</math> and <math>CFE</math> be <math>Ga, Gb,</math> and <math>Gc,</math> respectively.
+Let <math>Pa</math> be a point such that <math>EPa</math> is parallel to <math>CP</math> and <math>DPa</math> is parallel to <math>BP.</math>
+Symilarly, <math>Pb:  DPb||AP, FPb||CP, Pc: FPc||BP, EPc||AP.</math>
+Let <math>\triangle A'B'C'</math> be the triangle formed by the lines <math>GaPa, GbPb,</math> and <math>GcPc.</math>
+Let the point <math>X</math> be the centroid of the set of points <math>A, B, C, D, E, F.</math>
+Then <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center is the point <math>X,</math> the points <math>P, G,</math> and <math>G'</math> are collinear.
+One can prove this claim using the method of previous section.
+The points <math>P, Pa, Pb,</math> and <math>Pc</math> are collinear.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+== Dao Thanh Oai's Generalization==
+[[File:Generalization.png|400px|right]]
+Let triangle <math>ABC</math> and line <math>\ell,</math> non parallel to sidelines be given. Let line <math>\ell</math> meets sidelines <math>AB, AC, BC</math> of <math>\triangle ABC</math> at points <math>D, E, F,</math> respectively.
+Let <math>Q \in \ell.</math> Let <math>K, L, M</math> be the points such that <math>QA||EM||DL, QB||KD||FM, QC||KE||FL.</math>
+Similarly define points <math>P \in \ell, K_0, L_0, M_0.</math>
+Let triangle <math>\triangle A'B'C'</math> be the triangle formed by the lines <math>KK_0, LL_0, MM_0.</math>
+Prove that <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center lies on <math>\ell.</math>
+<i><b>Proof</b></i>
+[[File:Dao Generalization.png|400px|right]]
+Let <math>G</math> and <math>G_0</math> be the midpoints <math>BE</math> and <math>CD,</math> respectively.
+Let <math>X</math> be the crosspoint of <math>\ell</math> and Gauss line <math>GG_0.</math>
+Let <math>A', B', C', D', E',</math> and <math>K'</math> be the points simmetric to <math>A, B, C, D, E,</math> and <math>K</math> with respect to <math>X,</math> respectively.
+We will prove that <math>K' \in BC</math> which means that   <math>K \in B'C' \implies K_0 \in BC, L \in AC</math> and so on.
+<math>G</math> midpoint <math>BE, X</math> midpoint <math>EE' \implies GX||BE'.</math>
+<math>G_0</math> midpoint <math>DC, X</math> midpoint <math>DD' \implies G_0X||CD' \implies BE'||CD'.</math>
+<math>KD||K'D'||BQ, KE||K'E'||CQ, CD'||BE', Q \in D'E' \implies K' \in BC</math> according the Claim.
+<i><b>Claim (Parallel lines in trapezium)</b></i>
+[[File:Pappus.png|300px|right]]
+[[File:Pappus non convex.png|300px|right]]
+Let <math>ABCD</math> be the quadrungle such that <math>AD||BC.</math>
+Let <math>M \in AB, MD||BN, AN||CM.</math>
+Prove that point <math>N</math> lyes on <math>CD.</math>
+<i><b>Proof</b></i>
+We prove Claim in the case <math>AB</math> non parallel to <math>CD.</math>
+Denote <math>Q = AB \cap CD.</math>
+<math>AD||BC \implies \frac {QB}{QA}= \frac{QC}{QD}.</math>
+Let <math>BN||MD</math> cross <math>CD</math> at <math>N.</math> Then <math>\frac {QN}{QD}=\frac {QB}{QM}.</math>
+<cmath>\frac {QC}{QN} = \frac {QC}{QD} \cdot  \frac {QD}{QN} =  \frac {QB}{QA} \cdot  \frac {QM}{QB} = \frac {QM}{QA} \implies AN||CM.</cmath>
+The Claim is correct in the case of non convex <math>ABCD.</math> One can simplify the proof of  Dao Generalization using this variant of the Claim.
 '''vladimir.shelomovskii@gmail.com, vvsss'''