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My question: I am looking for a proof of problem as following:

Introduction: When I research a theorem as following:

Theorem 1: Let $ABC$ be a triangle, let $(S)$ be a circumconic of $ABC$, let $P$ be a point on the plane. Let the lines $AP, BP, CP$ meet the conic again at $A', B', C'$. Let $D$ be a point on the polar of point $P$ with respect to $(S)$ or $D$ lies on the conic $(S)$. Let $A_0= DA' \cap BC$, $B_0= DB' \cap AC$; $C_0= DC' \cap AB$. Then $A_0, B_0, C_0$ are collinear. Further more four points $A_0, B_0, C_0, P$ are collinear if only if $D$ lie on the conic.

1-Nguyen Ngoc Giang, A proof of Dao theorem, Global Journal of Advanced Research on Classical and Modern Geometries, ISSN: 2284-5569, Vol.4, (2015), Issue 2, page 102-105

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I found a nice result adding to configuration of theorem 1 as following:

Problem: In the case D lies on the conic $(S)$. Let $AP$ meets $BC$ at $P_a$. Let the line through $D$ and parallel to $AP$, this line meets $BC$ at $D_a$. Let $D'_a$ on the ray $DD_a$ such that $\frac{\overline{DD_a}}{\overline{DD'_a}}=\frac{\overline{A'P_a}}{\overline{A'P}}$. Define $D'_b, D'_c$ cyclically. Then seven points $A_0, B_0, C_0, D'_a, D'_b, D'_c$ and $P$ are collinear.

2-The problem is the problem 11 in this paper

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  • $\begingroup$ what is "nine points conic of P"? it would be more clear if you formulated this in terms of equations, than in geometric terms $\endgroup$ Mar 26, 2016 at 12:06
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    $\begingroup$ I thank to You, please see : en.wikipedia.org/wiki/Nine-point_conic $\endgroup$ Mar 26, 2016 at 12:09
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    $\begingroup$ If you apply an affine transform and move (ABC, P) to a triangle and orthocenter will you get the regular Feuerbach theorem? $\endgroup$ Mar 26, 2016 at 23:42
  • $\begingroup$ I thank to You very much dear @akopyan $\endgroup$ Mar 27, 2016 at 4:24
  • $\begingroup$ I deleted old result. I keep new result. $\endgroup$ Aug 3, 2016 at 17:28

1 Answer 1

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If I understand the problem correctly, it is already known that $A_0,B_0,C_0$ and $P$ are collinear as stated in Theorem 1 above or proven here. Call this common line $L$. It remains to prove that $D_a',D_b'$ and $D_c'$ lie on $L$. For this, suppose that the ray $DD_a$ intersects $L$ at $D_a''$. Using similarities $\Delta A_0DD_a\sim \Delta A_0A'P_a$ and $\Delta A_0DD_a''\sim \Delta A_0A'P,$ one has $$\frac{\overline{DD_a}}{\overline{A'P_a}}=\frac{\overline{A_0D}}{\overline{A_0A'}}=\frac{\overline{DD_a''}}{\overline{A'P}}\Rightarrow\frac{\overline{DD_a}}{\overline{DD_a''}}=\frac{\overline{A'P_a}}{\overline{A'P}},$$which shows that $D_a''$ coincides with $D_a'$. This shows that $D_a'$ lies on $L$. Similarly (repeating the argument cyclically), $D_b'$ and $D_c'$ lie on $L$. Hence the seven points $A_0,B_0,C_0,D_a',D_b',D_c'$ and $P$ are collinear.

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  • $\begingroup$ please check your email, or your junk email $\endgroup$ Feb 11, 2019 at 2:23

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