View topic - A difficult problem • Art of Problem Solving

Very nice problem, now here is my synthetic solution:

Denote $X \in EF \cap CG$ , and $T \in EF \cap BC$ .

Now because the quadrupoint $(T,B,D,C)$ forms an harmonical division, i.e $(T,B,D,C)=-1$ , so does the pencil $F(T,B,D,C)$ , and now by intersecting it with $CG$ , we obtain that $(X,G,N,C)=-1$ . $(1)$ .

Now on other hand by applying the Menelaus theorem in $\triangle{BCG}$ , for the transversal $\overarrow{D-N-F}$ , we obtain that $CD=3GF \iff CN=3NG$ .

But from $(1)$ , we have that $\frac{CN}{NG}=\frac{CX}{GX}$ .

Therefore it is sufficient to prove that $N$ is the midpoint of $(CX)$ , from there trivially resulting that $\frac{CX}{NX}=2=\frac{GX}{NX}\Rightarrow CD=3GF$ .

Now observe that $\angle{MEX}=\angle{MDF}=\angle{MCX}$ , so the quadrilateral $MECX$ is cyclic, so $\angle{MXC}=\angle{MEA}=\angle{ADE}$ . $(2)$

And $\angle{MCX}=\angle{ADF}$ . $(2')$

Again, $\angle{CMN}=\angle{FDB}$ $(3)$ , and $\angle{XMN}=\angle{XMC}-\angle{CMN}=\angle{CEF}-\angle{FDB}=\angle{EDC}$ . $(3')$ .

Now using $(2)$ , $(2')$ , $(3)$ , $(3')$ in $\frac{NX}{NC}=\frac{\sin{MCX}}{\sin{MXC}}\cdot \frac{\sin{XMN}}{\sin{CMN}}$ we obtain that:

$N$ is the midpoint of $(CX) \iff \frac{\sin{FDA}}{\sin{EDA}}= \frac{\sin{BDF}}{\sin{CDE}}$

But because $DA$ is a symmedian in $\triangle{EDF}$ , we have that $\frac{\sin{FDA}}{\sin{EDA}}=\frac{FD}{ED}=\frac{\sin{DEF}}{\sin{DFE}}=\frac{\sin{BDF}}{\sin{CDE}}$ .
Therefore $N$ is the midpoint of $(CX)$ , so the problem is solved.

A difficult problem

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