57-A generalization Newton's line theorem

Let the line (d) meets three sides of the triangle at $A_1,B_1,C_1$. Denote $A_2,B_2,C_2$ are midpoints of $AA_1,BB_1,CC_1$ then $A_2,B_2,C_2$ are collinear. The line through $A_2,B_2,C_2$ call Newton's line
 
Theorem(A generalization Newton line theorem): Let $P$ be a point on the plain,$EFG$ be cevian triangle of the point P. The line AA_1,BB_1,CC_1 meets three sidelines of $EFG$ at $A'_2, B'_2,C'_2$ then $A'_2, B'_2,C'_2$ are collinear. 

-  When $P$ is the centroid of $ABC$, this theorem is Newton line theorem above

 Solution by Telv Cohl (Telv Cohl: National Chiayi Senior High School, Chiayi, Taiwan. E-mail address: telvcohltinasprout@gmail.com)
Proof 1 :
$X$ = the intersection of $AE$ and $GF$
$Y$ = the intersection of $BF$ and $GE$
$Z$ = the intersection of $CG$ and $EF$

By Ceva theorem we get $\frac{AG}{GB}.\frac{BE}{EC}.\frac{CF}{FA}=1$                                                      (1)

By Menelaus theorem we get $\frac{CA_1}{A_1B}.\frac{BC_1}{C_1A}.\frac{AB_1}{B_1C}=1$                        (2)

By Ceva theorem we get $\frac{GX}{XF}.\frac{FZ}{ZE}.\frac{EY}{YG}=1$                                                      (3)

Since $(B,C;E,A_1)=(G,F;X,A_2)$ we get $\frac{BE}{EC}.\frac{CA_1}{BA_1}=\frac{GX}{XF}.\frac{FA_2}{A_2G}$         (4)

Since $(C,A;F,B_1)=(E,G;Y.B_2)$ we get $\frac{CF}{FA}.\frac{CB_1}{AB_1}=\frac{EY}{YG}\frac{GB_2}{B_2F}$             (5)

Since $(A,B;G,C_1)=(F,E;Z,C_2)$ we get $\frac{AG}{GB}.\frac{BC_1}{AC_1}=\frac{FZ}{ZE}.\frac{EC_2}{C_2F}$           (6)

From (1) (2) (3) (4) (5) (6) we get $\frac{FA_2}{A_2G}.\frac{GB_2}{B_2E}.\frac{EC_2}{C_2F}=1$

By the converse of Menelaus theorem we get $A_2, B_2, C_2$ are collinear
Q.E.D

56-Two reflection theorem

Problem 1: Let ABC be a triangle, let A’,B’,C’ such that AA’, BB’,CC’ are parallel and midpoint of AA’,BB’,CC’ lie on a line (d). Let P lie on (d), Denote three lines PA’,PB’,PC’ meet three side lines BC, CA, AB respectively at A1,B1,C1. Show that A1,B1,C1 are colliner.

Problem 2: Let A’B’C’ are reflection of a triangle ABC in line (d). Let P lie on (d). Let three line (d_a),(d_b),(d_c) through P and perpendicular with PA,PB,PC respectively, wich lines meet three side lines B’C’, C’A’, A’B’ respectively at A1,B1,C1. Show that A1,B1,C1 are colliner.

55-Three point collinear again

Let ABC be quadrilateral, M, N are midpoints of AD,BC respectively. Three red parallel lines through B,M,C respectively meet three blue parallel lines through A,ND respectively at I,J,K. Show that I,J,K are collinear and the line IJK through fixed point when three parrallel line turn. Solution click:

Two Pencils of Parallel Lines

54-A generalization Colling theorem

P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes, 2001; with corrections, 2013, available athttp://math.fau.edu/Yiu/Geometry.html at 2.4.3 More on reflections

(1) (Colling) The reflections of a line L in the side lines of triangle ABC are concurrent if and only if L passes through the orthocenter. In this case, the intersection is a point on the circumcircle.

I generalization this result following:

Let  be a triangle,  is the orthocenter of the triangle . are projection of  to . Let  lie on such that: . Let  be any point on the plain.  are reflection of  on . Show that:  are concurrent. If  we have Colling's theorem above

Proof by Telv Cohl https://www.facebook.com/telv.cohl?fref=ufi:

Lemma:
Given triangle ABC and a point P
D is the projection of A on BC 
E is the projection of B on CA 
F is the projection of C on AB 
Pa Pb Pc is the reflection of P wrt BC CA AB
prove that PaD PbE PcF are concurrent

Proof lemma:
Since PaD is the reflection of PD wrt AD so PaD is the isogonal line of PD wrt angle FDE similarity. PbE is the isogonal line of PE wrt angle DEF. PcF is the isogonal line of PF wrt angle EFD. Hence PaD PbE PcF are concur at the isogonal conjugate of P wrt triangle DEF done

Proof theorem:

Let H be the orthocenter of triangle ABC
S1 be the intersection of A1Da and B1Db
S2 be the intersection of B1Db and C1Dc
D' be the isogonal conjugate of D wrt triangle HaHbHc
K be the pole of the simson line which is parallel to DH

We move A1 on AH and move B1 on BH accordingly so as to preserve the condition given in hypothesis the map A1 --- > B1 is homography

so the map pencil{DaA1} --- > pencil{DbB1} is homography Hence the locus of S1 is a conic through Da and Db ... (1)

When A1=A2=H, S1 coincide with H ... (2)

When A1=infinity point of the direction perpendicular to BC then B1=infinity point of the direction perpendicular to CA S1 coincide with P ... (3)

When A1=Ha B1=Hb from lemma we deduce that S1 coincide with D' ... (4)

When A1=the intersection of AH and the circumcircle of triangle ABC
then B1=the intersection of BH and the circumcircle of triangle ABC
S1 concide with K ... (5)

From (1) (2) (3) (4) (5)
we get the locus of S1 is the conic through Da Db H P D' K ... (6)
similarity we get the locus of S2 is the conic through Db Dc H P D' K ... (7)

From (6) (7) we get the locus of S1 is coincide with the locus of S2
ie. S1 is coincide with S2 so DaA1 DbB1 DcC1 concur at one point
Q.E.D

53-A generalization of many theorem

Please see the file attachments and please see online at:  HechosGeometricos

viernes, 29 de noviembre del 2013 or (ADGEOM #893, Dao Thanh Oai )

This result is a generalization Jerabek hyperbola, Kiepert hyperbola, Feuerbach hyperbola, Kariya theorem, Goormaghtigh(page 17 at Forumgeometricorum) theorem, orthotransversal line theorem and many many another theorem.  if you like this result please add this result into your book.  Click Download

52-On the two points X(5463),x(5464)

X(13),X(15),X(5463),X(5464) lie on a circle.

X(14),X(16),X(5463),X(5464) lie on a circle.

X(2),X(110),X(5463),X(5464) lie on a circle
X(3),X(15),X(110),X(5463) lie on a circle
X(3),X(16),X(110),X(5464) lie on a circle
Circle through X(13),X(16),X(5464) and X(13)X(14)X(16) tangent with the line X(14)X(5464)
Circle through X(13),X(16),X(5464) and X(13)X(14)X(16) tangent with the line X(14)X(5464)
Circle through X(14),X(15),X(5463) and X(13)X(14)X(15) tangent with the line X(13)X(5463)

51-Many new circle in a triangle

Problem 1:

a- Euler line tangent with the circle through (X(381),X(98),X(597)) at X(381)
b- Euler line tangent with the circle through (X(3),X(98),X(597)) at X(3)
Since a,b we have if I,X(381),J, (X3) are harmonic range then I,J,X(98),X(597) lie on a circle

Problem 2: X(74)= intersection of Jerabek hyperbola with circumcircle. The line though X(74) and parallel with X(13)X(14) meets circumcircle again, meets Euler line, meest Brocard axis at M,Q,P respectively. T is reflection of X(3) on Q. Then we have problem
a- Euler line tangent with the circle through X(3),M,P
b- Euler line tangent with the circle through Q,M,P
Since a,b we have if I,X(3),J,T are harmonic range then I,J,M,P lie on a circle

Problem 3: X(74)= intersection of Jerabek hyperbola with circumcircle. The line though X(74) and parallel with X(13)X(14) meets Euler line, Brocard axis at Q,P respectively. T is reflection of X(3) on P. Then we have problem
a- Brocard axis tangent with the circle through X(74),X3,Q at X(3)
b- Brocard axis tangent with the circle through X(74),Q,T at T
Since a,b we have if I,X(3),J,T are harmonic range then I,J,X(74),Q lie on a circle

Problem 4: The line X(2)X(110) meets Jerabek hyperbolar at D,E then tangent of Jerabek hyperbola at D,E parallel with Brocard axis.
a-The circle through D,E,X(3) tangent with Brocard axis at X(3)
b- Circle through D,E,X(6) tangent with Brocard axis at X(6)
c- D,E, lie on a circke, special D,E,X(15),X(16) lie on a circle.

Problem 5: The line X(2)X(110) meets Jerabek hyperbolar at D,E,X(13),X(14) lie on a circle

Problem 6: Let ABC be a triangle, Let a rectangular circumhyperbola, Center is J. The line is isogonal conjugate of the hyperbolar meets the hyperbolar again at D,E. The hyperbolar meets circle again at M. (d) is a line through M and parallel with EJ. d meets the hyperbolar and the circumcircle at N,P. Then D,E,N,P lie on a circle

Problem 7: 
Xn is reflection of X(1) on X(11) . Then circle through X(1),X(n),X(4) tangent with the line through X(4)X(8) at X(4). Then circle through X(1),X(n),X(8) tangent with the line through X(4)X(8) at X(8)

50-Two circle, parallel line and collinear

1- is a cyclic quadrilateral,  is a cyclic quadrilateral.  meets  at .  meets  at .  meets  at . Then  are collinear

2-AB meets EC at G; CF meets BD at I; AF meets ED at H; AE//DF. Then G,H,I are collinear

49-Concurrent again be a point lie on Nine point circle

Let  be a triangle,  is the orthocenter.  be any point on the plane. Nine point circles of  concurrent at midpoint of . Them also concurrent again be a point lie on Nine point circle of the triangle 

48-Euler line of triangle bounded by three reflection parallel line

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C respectively. The reflections of L1,L2,L3 in BC,CA,AB, resp. bound a triangle A1B1C1. The Euler line of A1B1C1 passes through a fixed point (as the three lines L1,L2,L3 move around A,B,C, being parallel)

47-Pair triangle homothetic

Let two triangle  and . Such that ; .  meet  at . Define cyclically. Denote  is center of circumcircle  and . Denote Oa is center of circle , Define  cyclically.  is center of . Define  cyclically.
Prove that: 

1- tangent with three circle 
2- are concurrent at  and  is midpoints of ; when are median triangle then  is midpoints of center Nine point circle and Circumcircle

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