Jutaro wrote: Let $ ABCD$ be a convex quadrilateral and $ I = AC\cap BD$ , $ E\in (AB)$ , $ H\in (BC)$ , $ F\in (CD)$ , $ G\in (DA)$ such that $ I\in EF \cap GH$ . Denote $ M\in EG \cap AC$, $ N\in HF \cap AC$ . Prove that $ \frac {MA}{MI}\cdot \frac {NI}{NC} = \frac {IA}{IC}$ . Lemma 1. Let $ ABC$ be a triangle and $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ and $ P\in AD\cap EF$ . Then $ \frac {PD}{PA}\cdot BC = \frac {EC}{EA}\cdot DB + \frac {FB}{FA}\cdot DC$ .

Lemma 2. Let $ ABCD$ be a convex quadrilateral and $ I\in AC\cap CD$ , $ E\in (AB)$ , $ F\in (CD)$ such that $ I\in EF$ . Then $ \frac {EB}{EA}\cdot\frac {FC}{FD} = \frac {IB}{ID}\cdot\frac {IC}{IA}$ .

Proof of the proposed problem. $ \left\|\begin{array}{ccccc} I\in HG & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {HB}{HC}\cdot\frac {GA}{GD} = \frac {IB}{ID}\cdot\frac {IA}{IC} & \implies & \frac {GD}{GA}\cdot IB = \frac {HB}{HC}\cdot\frac {IC}{IA}\cdot ID \\ \\ I\in EF & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {EB}{EA}\cdot \frac {FC}{FD} = \frac {IB}{ID}\cdot\frac {IC}{IA} & \implies & \frac {EB}{EA}\cdot ID = \frac {FD}{FC}\cdot\frac {IC}{IA}\cdot IB\end{array}\right\|$ $ \bigoplus\ \implies$ $ \frac {GD}{GA}\cdot IB + \frac {EB}{EA}\cdot ID = \frac {IC}{IA}\cdot\left(\frac {HB}{HC}\cdot ID + \frac {FD}{FC}\cdot IB\right)$ $ \stackrel{(lemma\ 1)}{\ \ \implies\ \ }$ $ \frac {MI}{MA}\cdot BD = \frac {IC}{IA}\cdot\left(\frac {NI}{NC}\cdot BD\right)$ $ \implies$ $ \boxed {\ \frac {MA}{MI}\cdot\frac {NI}{NC} = \frac {IA}{IC}\ }$ .

Dedicate to Armpist ... Jutaro wrote: Let $ ABCD$ be a convex quadrilateral and $ I = AC\cap BD$ , $ E\in (AB)$ , $ H\in (BC)$ , $ F\in (CD)$ , $ G\in (DA)$ such that $ I\in EF \cap GH$ . Denote $ M\in EG \cap AC$, $ N\in HF \cap AC$ . Prove that $ \frac {MA}{MI}\cdot \frac {NI}{NC} = \frac {IA}{IC}$ . Lemma 1. Let $ ABC$ be a triangle and $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ and $ P\in AD\cap EF$ . Then $ \frac {PD}{PA}\cdot BC = \frac {EC}{EA}\cdot DB + \frac {FB}{FA}\cdot DC$ .

Lemma 2. Let $ ABCD$ be a convex quadrilateral and $ \underline{I\in AC\cap BD}$ , $ E\in (AB)$ , $ F\in (CD)$ such that $ I\in EF$ . Then $ \frac {EB}{EA}\cdot\frac {FC}{FD} = \frac {IB}{ID}\cdot\frac {IC}{IA}$ .

Proof of the proposed problem. $ \left\|\begin{array}{ccccc} I\in HG & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {HB}{HC}\cdot\frac {GA}{GD} = \frac {IB}{ID}\cdot\frac {IA}{IC} & \implies & \frac {GD}{GA}\cdot IB = \frac {HB}{HC}\cdot\frac {IC}{IA}\cdot ID \\ \\ I\in EF & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {EB}{EA}\cdot \frac {FC}{FD} = \frac {IB}{ID}\cdot\frac {IC}{IA} & \implies & \frac {EB}{EA}\cdot ID = \frac {FD}{FC}\cdot\frac {IC}{IA}\cdot IB\end{array}\right\|$ $ \bigoplus\ \implies$ $ \frac {GD}{GA}\cdot IB + \frac {EB}{EA}\cdot ID = \frac {IC}{IA}\cdot\left(\frac {HB}{HC}\cdot ID + \frac {FD}{FC}\cdot IB\right)$ $ \stackrel{(lemma\ 1)}{\ \ \implies\ \ }$ $ \frac {MI}{MA}\cdot BD = \frac {IC}{IA}\cdot\left(\frac {NI}{NC}\cdot BD\right)$ $ \implies$ $ \boxed {\ \frac {MA}{MI}\cdot\frac {NI}{NC} = \frac {IA}{IC}\ }$ .

Notice that triangles $ \triangle AEG$ and $ \triangle CFH$ are perspective through $I,$ thus by Desargues theorem, the intersections $ X \equiv AD \cap BC,$ $ Y \equiv AB \cap DC$ and $ P \equiv EG \cap HF$ are collinear. Let $ Q \equiv AC \cap XY.$ Then by Menelaus' theorem for $ \triangle MQP$ and $ \triangle NQP$ cut by the straight lines $ AB$ and $ CD,$ we have $ \frac {YQ}{YP} \cdot \frac {PE}{EM} \cdot \frac {MA}{QA} = 1 \ \ , \ \ \frac {YQ}{YP} \cdot \frac {PF}{NF} \cdot \frac {NC}{QC} = 1$ $ \Longrightarrow \frac {MA}{NC} \cdot \frac {PE}{EM} \cdot \frac {NF}{PF} = \frac {QA}{QC} \ (\star)$ But by Menelaus' theorem for $ \triangle MNP$ cut by $ \overline{FIE},$ we have $\frac {PE}{EM} \cdot \frac {NF}{PF} = \frac {IN}{IM}$ Combining the latter expression with $ (\star)$ yields $ \frac {MA}{NC} \cdot \frac {IN}{IM} = \frac {QA}{QC}$ Since $ (A,C,I,Q) = - 1,$ it follows that $ \frac {AM}{IM} \cdot \frac {IN}{CN} = \frac {QA}{QC} = \frac {IA}{IC}.$

we can use projective transformation

Note that the problem statement is purely projective. Also, it is well-known that there is a projective transformation which maps $ABCD$ to a square $A'B'C'D'$. Let this transformation map $E,F,G,H,I,M,N$ to $E',F',G',H',I',M',N'$, respectively. [asy][asy] defaultpen(fontsize(10pt)); size(150); pair A=dir(135); pair B=dir(225); pair C=dir(315); pair D=dir(45); pair I=extension(A,C,B,D); pair E=0.6*B+0.4*A; pair H=0.8*B+0.2*C; pair F=extension(E,I,C,D); pair G=extension(H,I,A,D); pair M=extension(E,G,A,C); pair N=extension(H,F,A,C); dot("$A'$", A, dir(A)); dot("$B'$", B, dir(B)); dot("$C'$", C, dir(C)); dot("$D'$", D, dir(D)); dot("$E'$", E, dir(E)); dot("$F'$", F, dir(F)); dot("$G'$", G, dir(G)); dot("$H'$", H, dir(H)); dot("$I'$", I, dir(270)); dot("$M'$", M, dir(185)); dot("$N'$", N, dir(0)); draw(A--B--C--D--A--cycle); draw(E--G--H--F--E--cycle); draw(A--C); draw(B--D); [/asy][/asy] As projective transformations preserve cross-ratio and thanks to the central symmetry in a square, we have that \begin{align*} (A,I;M,C)&=(A',I';M',C')\\ &=\dfrac{\frac{A'M'}{M'I'}}{\frac{A'C'}{C'I'}}\\ &=\dfrac{\frac{C'N'}{N'I'}}{\frac{C'A'}{A'I'}}\\ &=(C',I';N',A')\\ &=(C,I;N,A)\\ \end{align*} Finally, this implies that \begin{align*} \dfrac{\frac{AM}{MI}}{\frac{AC}{CI}}&=\dfrac{\frac{CN}{NI}}{\frac{CA}{AI}}\\ \dfrac{AM}{IM}\cdot\dfrac{IN}{CN}&=\dfrac{IA}{IC}\\ \end{align*} We are done. $\square$