79-On the Pascal line theorem

Cite Course: Cut The Knot

78-Cevian triangle associated with Jacobi Configuration

Let ABC be a triangle, let $\triangle J_aJ_bJ_c$, $\triangle J'_aJ'bJ'_c$ be two Jacobi triangle, $A_0=J_aJ'_a \cap BC$, define $B_0,C_0$ cyclically. Then $AA_0,BB_0,CC_0$ are concurrent

77-Transfomation line to line on the Pascal line

Let $A,B,C,C_1,B_1A_1$ lies on a conic. Let $A_2=BC_1 \cap B_1C, B_2=AC_1 \cap A_1C; C_2=AB_1 \cap A_1B$. Let $A_3,B_3,C_3$ lie on sidelines $BC,CA,AB$ such that $A_3,B_3,C_3$ are collinear. Let $A_4=A_3A_2 \cap B_1C_1, B_4=B_3B_2 \cap A_1C_1; C_4=A_3B_3 \cap A_1B_1$ . Then $A_4,B_4,C_4$ are collinear.

76-A generalization of Simson line theorem

Problem 1: Let $ABC$ be a triangle, let a line $(L)$ through circumecenter and a point $P$ lie on circumcircle.

Let $AP,BP,CP$ meets $(L)$ at $A_P, B_P, C_P$. 

Denote $A_0,B_0,C_0$ are projection (mean perpendicular foot) of $A_P, B_P, C_P$ to $BC,CA,AB$ respectively.

Then $A_0,B_0,C_0$ are collinear.

- When $(L)$ through $P$, this line is Simson line.

Problem 2: The new line $\overline {A_0B_0C_0}$ bisect the orthocenter and P

75-Two similar rectangles theorem

If ABCD and $A_1B_1C_1D_1$ are two similar rectangles (mean $AB:AD=A_1B_1:A_1D_1)$, then 

Red=Yellow <=> $area(AA_1B_1B)+area(CC_1D_1D)=area(AA_1D_1D)+area(CC_1B_1B)$

I found this result when I research the British flag theorem.

74-A circle tangent to circumcircle(Similar with the third Fontené)

Colling theorem: Let a line $(L)$ through the orthocener of the triangle ABC, reflection of $(L)$ in three sidelines concurrent be a point on the circumcircle(Colling point of $(L)$

Nice problem: Let $ABC$ be a triangle, let a line $(L)$ through the orthocenter of the triangle $ABC$ and $(L)$ cuts $AB, AC$ at $P, Q$. Denote $K$ be the intersection of two lines perpendicular to $AB,AC$ at $P,Q$. These line cut $BC$ at $M,N$. The circle through $KMN$ tangent the circumcircle at colling point of (L).

Note: $X_3$ of $ABC$ is $X_4$ of median triangle

73-X(4240) =12 EULER LINES POINT

X(4240) in Kimberling Center

Click to downloand detail defined and the figure

Barycentrics   f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (b⁴ + c⁴ - 2a⁴ + a²b² + a²c² - 2b²c²)/[(b² - c²)(b² + c² - a²)]

X(4240) is the point of intersection of the Euler lines of nine triangles, constructed as in the next three paragraphs.

Let E be the Euler line of a triangle ABC. Let A₁ = E∩BC, and define B₁ and C₁ cyclically. Let A_B be the reflection of A in B₁, and define B_C and C_A cyclically. Let A_C be the reflection of C in B₁, and define B_A and C_B cyclically. The Euler lines of the four triangles ABC, AA_BA_C, BB_CB_A, CC_AC_B concur in X(4240). (Dao Thanh Oai, Problem 1 in attachment to ADGEOM #1709, September 15, 2014). See also Telv Cohl, 'Dao's Theorem on the Concurrency of Three Euler Lines,' International Journal of Geometry 3 (2014) 70-73.

Continuing, let A*B*C* be the paralogic triangle of ABC whose perspectrix is E. Then X(4240) lies on the Euler line of A*B*C*. (Dao Thanh Oai, noted just after Figure 1 in attachment to ADGEOM #1709, September 15, 2014).

Continuing, redefine A_B as the point on line AC and A_C as the point on line AB such that B₁, A₁, A_B, A_C line on a circle and A₁, A_B, A_C are collinear. Define B_C and B_A cyclically, and define C_A and C_B cyclically. Let A₂ = B_AB_C∩C_AC_B and define B₂ and C₂ cyclically. The Euler lines of the five triangles ABC, A₂B₂C₂, AA_BA_C, BB_CB_A, CC_AC_B concur in X(4240). (Dao Thanh Oai, Problem 2 in attachment to ADGEOM #1709, September 15, 2014).

X(4240) lies on these lines:
{2, 3}, {107, 110}, {112, 1302}, {476, 1304}, {925, 1301}, {2407, 3233

72-Concurrency on the Tucker circle

Let a Tucker circle cut sidelines of given triangle $ABC$ at $A_b,A_c, B_c,B_a,C_a,C_b$. If $A_2A_bA_c, B_2B_aB_c, C_2C_aC_b$ are three similar isosceles triangles then $AA_2BB_2CC_2$ are concurrent

71-A line through H

Let $ABC$ be a triangle, $B_a,C_a$ lie on $BC; C_b,B_c$ lie on $CA,BA$ respectively. Such that two circles $(BB_aB_c),(CC_aC_b)$ tangent at $P$ on the circumcircle and $B_cB_a \perp AC, C_aC_b \perp AB$. Denote $A_b=B_aB_c \cap AC, A_c=C_aC_b \cap AB$. Show that: $A_bA_c$ through the orthoenter of the triangle $ABC$. $M=B_cC_b \cap AO$, $M$ lie on circumcircle of $ABC$

70-Some problem on the Simson's line

Let $ABC$ be a triangle, let three points $A',B',C'$ on the circumcircle, such that $AA'//BB'//CC'$. Then the triangle $A_0B_0C_0$ form(bounded) by three Simson line of $A',B',C'$ 

1-$A_0B_0C_0$ are similar and perpective with $ABC$. Which is the locus of perpector? 

2- The circumcenter and the orthocenter $A_0B_0C_0$ , also lie on the circle with diameter $X(3)X(4)$ center $X(5)$

3-I don't know when the $A_0B_0C_0$ become to a point, but I known $A_0B_0C_0$ become a point three time when we moved $A',B',C'$ on circumcircle(such that $AA'//BB'//CC'$), these points also lie on the circle with diameter $X(3)X(4)$ center $X(5)$

69-New or old? Three lines concurrents

Let $ABC$ be a triangle, let three points $A_0,B_0,C_0$ lie on $BC,CA,AB$ such that $A_0,B_0,C_0$ are collinear. Let circle $(P)$ which center $P$ lie on the line $\overline{A_0B_0C_0}$. Denote $A_1,B_1,C_1$ are reflection of $A_0,B_0,C_0$ in $(P)$ show that $AA_1,BB_1,CC_1$ are concurrent.

68-Four centers are cyclic

- X(182) is midpoint of symmedian point X(6) and circumcenter X(3)

- X(98) is intersection of Kiepert hyperbola and circumcircle

Show that: X(98),X(182), The Nine point center and the Orthocenter lie on a circle

67-A generalization of Christipher Zeeman's theorem

Let $ABC$ be a triangle, let a line $(d)$ cut the sidelines $BC,CA,AB$ at $A_1,B_1,C_1,$ denote $H_A,H_B,H_C$ are the orthocenter of $AB_1C_1, BC_1A_1$, $CA_1B_1$ ; Denote $A',B',C'$ are reflection of $A$ in midpoint of $B_1C_1$; $B$ in midpoint of $C_1A_1$; C in midpoint of $A_1B_1$ respectively.

Theorem 1: Three line through $H_A,H_B,H_C$ and parallel to $BC,CA,AB$ form a triangle congruent and inversely homothety to $ABC$

Theorem 2: Three line through $A',B',C'$ and parallel to $BC,CA,AB$ form a triangle congruent and homothety to ABC, the homothety at infinity

66-A Generalization Gossard perspector and X(110)

Let $ABC$ be a triangle, let a line $(d)$ cut the sidelines $BC,CA,AB$ at $A_1,B_1,C_1$, denote $H_A,H_B,H_C$ are the orthocenter of $AB_1C_1$, $BC_1A_1$, $CA_1B_1$ ; Denote $A',B',C'$ are reflection of $A$ in midpoint of $B_1C_1$; $B$ in midpoint of $C_1A_1$; $C$ in midpoint of $A_1B_1$ respectively.

Theorem 1: Three line through $H_A,H_B,H_C$ and parallel to $BC,CA,AB$ form a triangle congruent and inversely homothety to $ABC$

Theorem 2: Three line through $A',B',C'$ and parallel to $BC,CA,AB$ form a triangle congruent and homothety to $ABC$, the homothety at infinity

65-A conjecture on polynomial function

Let a polynomial function: 
f(x)=$x^n+a_{n−1}x^{n−1}+⋯+a_2x^2+a_1x+a_0$
Where n is positive integer, $n \geq 3$ and $a_0, a_1, a_2, ..., a_n$ are constant coefficients   $\in Z$ and $f(x)=0$ has only solution $p \in Z$. Then has only $x=z,y =p or y=z,x =p or x=y=z=p$ with $x,y,z \in Z$ can satisfy the equation:

$f(x)+f(y)=f(z)$

64-A nice generalization Lester theorem

A Generalization Lester theorem follow: 

Theorem: Let two point  lie on one branch of a rectangular hyperbola. The line pass through midpoint of , and pass through center of the hyperbola cut hyperbola at two point . Let four points  are harmonic range then four point  lie on a circle.

63-Một tính chất quan trọng của hyperbol chữ nhật

Giới thiệu: Như ta đã biết luôn có một conic đi qua năm điểm phân biệt trên mặt phẳng. Nhưng với dấu hiệu nào ta có thể nhận ra đường conic là hyperbol,  ellipse, hoặc parabol? Trong bài viết này chỉ ra một dấu hiệu về tồn tại một Hyperbol chữ nhật qua năm điểm cho trước. Bài viết này là tổng quát của các vấn đề sau:

- Tổng quát của đường thẳng Droz-Farny,

- Tổng quát vấn đề 3878 trên tạp chí  Crux mathematicorum, issue 8, volum 39

- Tổng quát định lý về tương ứng vuông góc

Định lý 1: Cho $A,B,C,D,E$ nằm trên một đường hyperbol chữ nhật, ba đường thẳng qua $D$ và vuông góc với $EA,EB,EC$ giao với $BC,AC,AB$ lần lượt tại  $A_1,B_1,C_1$. Khi đó $A_1,B_1,C_1$ thẳng hàng và đường thẳng này vuông góc với $DE$.

                                         

Chứng minh:

Không giảm tổng quát, trong tọa độ Đề Các ta có thể giả sử đường hyperbol chữ nhật có phương trình là:

\begin{equation}y=\frac{p}{x} \end{equation}

(Lưu ý: Bất cứ đường hyperbol chữ nhật nào đều có thể đưa về dạng trên)

Và $A(a,\frac{p}{a})$, $B(b,\frac{p}{b})$, $C(c,\frac{p}{c})$, $D(d,\frac{p}{d})$, $E(e,\frac{p}{e})$. Phương trình đường thẳng qua  $D$ và vuông góc với $DA$ là:

\begin{equation}x-\frac{p}{a.e}y_{DA_1}=d-\frac{p^2}{e.a.d}\end{equation}

Tương tự phương trình đường thẳng $DB_1,DC_1$ là:

\begin{equation}x-\frac{p}{b.e}y_{DB_1}=d-\frac{p^2}{e.b.d}\end{equation}

\begin{equation}x-\frac{p}{c.e}y_{DA_1}=d-\frac{p^2}{e.c.d}\end{equation}

Phương trình đường thẳng  $BC,CA,AB$ lần lượt là:

\begin{equation}y_{BC}=-\frac{p}{c.b}x+\frac{p}{b}+\frac{p}{c}\end{equation}

\begin{equation}y_{CA}=-\frac{p}{c.a}x+\frac{p}{c}+\frac{p}{a}\end{equation}

\begin{equation}y_{BC}=-\frac{p}{a.b}x+\frac{p}{a}+\frac{b}{c}\end{equation}

Trong tọa độ điểm $A_1$ là nghiệm của phương trình (2) và (5).

$x-\frac{p}{a.e}y_{DA_1}=d-\frac{p^2}{e.a.d} \\ y_{BC}=-\frac{p}{c.b}x+\frac{p}{b}+\frac{p}{a}$  

Giải hệ phương trình tuyến tính hai ẩn ở trên ta có:

$x=\frac{d.c.p^2+d.b.p^2-c.b.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d} \\ y= \frac{e.a.c.p.d+e.a.b.p.d+p^3+e.a.p.d^2}{d.p^2+e.a.c.b.d}$ 

Do đó tọa độ điểm $A_1$ là:

$A_1(\frac{d.c.p^2+d.b.p^2-c.b.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d}, \frac{e.a.c.p.d+e.a.b.p.d+p^3-e.a.p.d^2}{d.p^2+e.a.c.b.d})$

Tương tự tọa độ các điểm $B_1,C_1$

$B_1(\frac{d.c.p^2+d.a.p^2-c.a.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d}, \frac{e.b.c.p.d+e.a.b.p.d+p^3-e.b.p.d^2}{d.p^2+e.a.c.b.d})$

$C_1(\frac{d.a.p^2+d.b.p^2-a.b.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d}, \frac{e.a.c.p.d+e.c.b.p.d+p^3-e.c.p.d^2}{d.p^2+e.a.c.b.d})$

$A_1,B_1,C_1$ thẳng hàng nếu và chỉ nếu: $\overrightarrow{A_1B_1}=k\overrightarrow{B_1C_1}$

$\overrightarrow{A_1B_1}=\frac{1}{d.p^2+e.a.c.b.d}(d.c.p^2+d.a.p^2-c.a.p^2+e.a.c.b.d^2-d.c.p^2-d.b.p^2 \\+ c.b.p^2-e.a.c.b.d^2 , e.b.c.p.d+e.a.b.p.d+p^3-e.b.p.d^2-e.a.c.p.d-e.a.b.p.d-p^3+e.a.p.d^2) \\ = \frac{1}{d.p^2+e.a.c.b.d}(d.a.p^2-c.a.p^2-d.b.p^2+c.b.p^2, e.b.c.p.d+e.b.p.d^2-e.a.c.p.d-c.a.p.d^2) \\ =\frac{(b-a)(c-d)p}{d.p^2+e.a.c.b.d}(p,ed)$

Tương tự như vậy chúng ta có:

$\overrightarrow{A_1C_1}=\frac{(c-a)(b-d)p}{d.p^2+e.a.c.b.d}(p, ed)$

$\Rightarrow$  \begin{equation}\overrightarrow{A_1B_1}= \frac{(b-a)(c-d)}{(c-a)(b-d)} \overrightarrow{A_1C_1}  \end{equation}

Do đo $A_1,B_1,C_1$ thẳng hàng. Hệ số góc của $DE$ là: $\frac{\frac{p}{e}-\frac{p}{d}}{e-d}=-\frac{p}{ed}$, hệ số góc của đường thẳng $\overline{A_1B_1C_1}$ là: $\frac{ed}{p}$. Do đó đường thẳng $\overline{A_1B_1C_1}$ vuông góc với đường thẳng $DE$. Định lý 1 chứng  hoàn tất.

Hệ quả 2 [1]:  Cho tam giác $ABC$, H là trực tâm, D là điểm bất kỳ trong mặt phẳng, ba đường thẳng qua $H$ và vuông góc với $DA,DB,DC$ giao với $BC,AC,AB$ lần lượt tại  $A_1,B_1,C_1$. Khi đó $A_1,B_1,C_1$ thẳng hàng và đường thẳng này vuông góc với $HD$.

Chứng minh:

Với điểm D bất kỳ thì A,B,C,H,D luôn nằm trên một hyperbol chữ nhật nên ta có điều phải chứng minh.

Hệ quả 3 [2]:  Cho tam giác $ABC$ và một điểm $P$ trên mặt phẳng, ba đường thẳng vuông góc với $PA,PB,PC$ tại $P$ lần lượt cắt ba cạnh $BC,CA,AB$ của tam giác tại $A_0,B_0,C_0$ thẳng hàng.

Chứng minh:

Với điểm P bất kỳ thì ta có 5 điểm A,B,C,P,P luôn nằm trên một hyperbol chữ nhật nên ta có điều phải chứng minh.

Định lý 4: Cho tam giác ABC, P,Q là hai điểm trên mặt phẳng sao 5 điểm A,B,C,P,Q nằm trền một hyperbol chữ nhật. Kẻ đường thẳng vuông góc AP qua Q, đường thẳng này cắt AB,AC tại $A_c,A_b$. Định nghĩa các điểm $B_a,B_c$,$ C_b,C_a$ tương tự. Khi đó sáu điểm $A_c,A_b, B_a,B_c$,$ C_b,C_a$ nằm trên một đường Conic.

Chứng minh:  Định lý 4 được chứng minh trực tiếp từ định lý 1 và định lý Pascal.

Tham khảo: 

[1]-Đào Thanh Oai, Vấn đề 3878 tạp chí Crux mathematicorum, issue 8, volum 39

[2]- Gibert, B. "Orthocorrespondence and Orthopivotal Cubics." Forum Geom. 3, 1-27, 2003. http://forumgeom.fau.edu/FG2003volume3/FG200301index.html.

Dao Thanh Oai: Kien Xuong, Thai Binh, Viet Nam.

Email address: daothanhoai@hotmail.com

62-Similar Goormaghtigh theorem, with Incenter

Problem: $I$ is incenter(or excenter), $A',B',C'$ lie on $IA,IB,IC$ such that $\frac{IA}{IA'}=\frac{IB}{IB'}=\frac{IC}{IC'}$ then three perpendicular of $IA,IB,IC$ at $A',B',C$ meets three sidelines at $A_1,B_1,C_1$ then $A_1,B_1,C_1$ are collinear.

Solution by: Telv Cohl https://www.facebook.com/telv.cohl

Let incircle of triangle $ABC$ touch $BC, CA, AB$ at $D, E, F$ and denote circle with diameter $RQ$ as circle $(RQ)$ 

Inverse with center $I$ and power $ID^2=IE^2=IF^2$ this inversion map $A, B, C, A', B', C'$ into $X, Y, Z ,X', Y', Z'$ and map $A_1, B_1 C_1$ into $A_1', B_1', C_1'$. Since $\frac{IX'}{IX}=\frac{IA}{IA'}=\frac{IB}{IB'}=\frac{IY'}{IY}$ we deduce that $X'Y' \parallel XY \parallel DE$, similarity we get $Y'Z' \parallel EF$ and $Z'X' \parallel FD$ so triangle $DEF$ and triangle $X'Y'Z'$ are homothety hence $DX', EY', FZ'$ concur at one point (denote this point as $P$) ...(1)

Notice that $A_1'$ is the intersection of circle $(IX')$ and circle $(ID)$ which is obvious the projection of $I$ on $DX'$. Similarity $B_1', C_1'$ is the projection of $I$ on $EY', FZ' $. From (1) we deduce that $I, A_1', B_1', C_1', P$ are concyclic at circle $(IP)$. So $A_1, B_1, C_1$ are collinear. 

Another solution by Leo mihai Giugiuc https://www.facebook.com/leo.giugiuc?fref=ts in the pictute

61-A property of incircle X(1)

Problem: Let $ABC$ be a triangle, $A_1B_1C_1$ be cevian triangle of the point $D$ (on the plane). I is the in center of the triangle ABC. Reflection of BC on IA1 meets $B_1C_1$ at $A_2$. Define $B_2,C_2$ cycllicaly. Show that $A_2,B_2,C_2$ are collinear.

                             

Solution by Telv Cohl https://www.facebook.com/telv.cohl

Let Incircle of triangle $ABC$ touch $BC, CA, AB$ at $G, E, F$. $L_g$ is the line through $G$ which is perpendicular to $IA_1$.  $L_e$ is the line through $E$ which is perpendicular to $IB_1$. $L_f$ is the line through $F$ which is perpendicular to $IC_1$. $L_g, L_e, L_f$ form triangle $A_3B_3C_3$
Let $G'$ be the reflection point of $G$ wrt $IA_1$, similarly define $E'$ and $F'$ Let $A_4$ be the intersection of $B_1C_1$ and $BC$, similarly define $B_4$ and $C_4$. ($A_4, B_4, C_4$ not appear in my picture)

Since $EE', FF'$ are the polar of $(B_1)$, $(C_1)$ wrt Incircle, So $B_1C_1$ is the polar of the intersection of $EE'$ and $FF'$ ie. $A_3$ similarly discussion we get $C_1A_1, A_1B_1$ is the polar of $(B_3), (C_3)$.

Since $AA_1, BB_1, CC_1$ concur at $D$ so triangle ABC and triangle A1B1C1 are perspective. By Desargue theorem we deduce that$A_4, B_4 C_4$ are collinear ... (1). Since $BC, CA, AB$ is the polar of $G, E, F$ wrt Incircle respectively. By Pole and Polar theorem and (1) we get $A_3G, B_3E, C_3F$ are concur. By Terquem theorem for triangle $A_3B_3C_3$ and Incircle we get $A_3G', B_3E', C_3F'$ are concur ... (2). Since $A_2, B_2, C_2$ is the pole of $A_3G', B_3E', C_3F'$ respectively. By Pole and Polar theorem and (2) we get $A_2, B_2,C_2$ are collinear. Q.E.D

60-A Problem of Seven circle

Problem:

Solution: by Telv Cohl 

Denote $O$ as the center of black circle, $O_{12}, O_{23}, O_{31}$ the center of three green circles $O_1, O_2, O_3$ the center of three red circles (in Dao's picture), $(O_1)$ tangent $(O_{12}) (O_{31})$ at $X, Y$ respectively. 

Let $H_1$ be the external center of similitude of $(O_{12})$ and $(O_{31})$ similarity define $H_2$ and $H_3$ inverse with center $H_1$ and power $H_1T_2.H_1T_3$. Since $H_1$ is the external center of similitude of $(O_{12})$ and $(O_{31})$ so this inversion map $(O_{12})$ and $(O_{31})$ to each other ... (1)

Easy to see that this inversion map $(O)$ to itself ... (2). Since $X$ is the internal center of similitude of $(O_{12})$ and $(O_1)$ and Y the internal center of similitude of $(O_1)$ and $(O_{31})$. 

By D'Alembert theorem we get $X, Y, H_1$ are collinear, and $H_1X.H_1Y=H_1T_2.H_1T_3$ hence this inversion also map $(O_1)$ to itself ... (3)

From (1) (2) (3) we deduce that this inversion map $(O_2)$ and $(O_3)$ to each other so $H_1, O_2, O_3$ are collinear  ie. $H_1$ is the intersection of $O_2O_3$ and $O_{12}O_{31}$ ... (4) similarity $H_2$ is the intersection of $O_3O_1$ and $O_{12}O_{23}$ ... (5) $H_3$ is the intersection of $O_1O_2$ and $O_{23}O_{31}$ ... (6)

By D'Alembert theorem we get $H_1, H_2, H_3$ are collinear ... (7) From (4) (5) (6) (7) we get $T_{12}T_{23}T_{31}$ and $T_3T_1T_2$ are perspective.
By Desargue theorem we get $T_1T_{23}, T_2T_{31}, T_3T_{12}$ are concurrent. Q.E.D

59-A generalization Goormaghtigh theorem and Zaslavsky's Theorem

The problem appeared in  O.T.Dao, Advanced Plane Geometry Message 1307 , May 22, 2014

58-Two nice line in a triangle

The first line: 

Let  be a triangle, a line  meets  at  are midpoints of .  are reflection of  in . Then  are collinear.

Solution:

http://www.cut-the-knot.org/pythagoras/AMatterOfCollinearity.shtml

The Secon line:

Let  be a triangle,  are projection of  on .  be a point on the plane.  lie on  respectively, define  cyclically such that .  is intersection of perpendicular of  at . Define  cyclically. Then  are collinear. Please see the figure attachments

Solution: by Telv Cohl https://www.facebook.com/telv.cohl?fref=ts

Let A4 B4 C4 be the mid point of GA3 GB3 GC3 and H' be the orthocenter of triangle A1B1C1. Easy to see GA3 is the diameter of circle(GA1A2A3)
and A4 is the center of circle(GA1A2A3) similarity for B3 B4 and C3 C4

Since A1H' x H'A2=B1H' x H'B2=C1H' x H'C2 so (GA1A2A3) (GB1B2B3) (GC1C2C3) are coxial ( with common radical axis GH' ) hence A4 B4 C4 are collinear ie.A3 B3 C3 are collinear

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