# A new theorem in projective geometry

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.

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.

• what is "nine points conic of P"? it would be more clear if you formulated this in terms of equations, than in geometric terms – Dima Pasechnik Mar 26 '16 at 12:06
• I thank to You, please see : en.wikipedia.org/wiki/Nine-point_conic – Oai Thanh Đào Mar 26 '16 at 12:09
• If you apply an affine transform and move (ABC, P) to a triangle and orthocenter will you get the regular Feuerbach theorem? – Arseniy Akopyan Mar 26 '16 at 23:42
• I thank to You very much dear @akopyan – Oai Thanh Đào Mar 27 '16 at 4:24
• I deleted old result. I keep new result. – Oai Thanh Đào Aug 3 '16 at 17:28

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.