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

enter image description here

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

enter image description here

  • $\begingroup$ 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
  • 1
    $\begingroup$ I thank to You, please see : en.wikipedia.org/wiki/Nine-point_conic $\endgroup$ Mar 26, 2016 at 12:09
  • 1
    $\begingroup$ 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


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.