Bumpp...

This is the rather well-known Poncelet's Theorem Let $\triangle A_1B_1C_1$ and $\triangle D_1E_1F_1$ be the contact triangles of $\triangle ABC$ and $\triangle DEF$, respectively in the common incircle $\omega$. Set $P_{XY}$ as the midpoint of $X_1Y_1$ where $X, Y \in \{A, B, C, D, E, F\}$. Lemma Points $A, B, C, D, E, F$ lie on a conic. (Proof) Note that $$C(DE, AB) \overset{C}{=} (D, E; CA \cap DE, CB \cap DE) \overset{I}{=} (P_{FD}, P_{FE}; P_{FB}, P_{FA}) \overset{F_1}{=} (D_1, E_1; B_1, A_1).$$It follows that $C(DE, AB)=F(DE, AB)$ and so the six points lie on a conic. $\square$ Finally, we remark that a non-degenerate conic is uniquely determined by five points on it, so $F$ lies on the circle passing through $A, B, C, D, E$ and we may conclude.

anantmudgal09 wrote: This is the rather well-known Poncelet's Theorem Let $\triangle A_1B_1C_1$ and $\triangle D_1E_1F_1$ be the contact triangles of $\triangle ABC$ and $\triangle DEF$, respectively in the common incircle $\omega$. Set $P_{XY}$ as the midpoint of $X_1Y_1$ where $X, Y \in \{A, B, C, D, E, F\}$. Lemma Points $A, B, C, D, E, F$ lie on a conic. (Proof) Note that $$C(DE, AB) \overset{C}{=} (D, E; CA \cap DE, CB \cap DE) \overset{I}{=} (P_{FD}, P_{FE}; P_{FB}, P_{FA}) \overset{F_1}{=} (D_1, E_1; B_1, A_1).$$It follows that $C(DE, AB)=F(DE, AB)$ and so the six points lie on a conic. $\square$ Finally, we remark that a non-degenerate conic is uniquely determined by five points on it, so $F$ lies on the circle passing through $A, B, C, D, E$ and we may conclude. A special case of it I must say. My solution was to just quote Poncelet's theorem and then prove it and use it's very very special case as the problem.

@above, they aren't that different. Use projective transformation to send the inconic to the incircle and proceed with the same proof as shown here

Yeah precisely. My solution used the Poncelet's theorem for circles and used $n=3$.

where can i get an article for it.