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$\triangle ABC$ has fixed line segment $BC$ and moving point $A$. $E$ and $F$ are points on the bisector of $\widehat{CAB}$ such that $\widehat{ABE} = \widehat{ACB}$ and $\widehat{ACF} = \widehat{ABC}$. Let $x$ and $y$ be lines that pass through respectively $B$ and $C$ and parallel to $EF$. $x \cap CA = M$ and $y \cap BA = N$. Prove that the intersection of $ME$ and $NF$ is a fixed point.

Let $ME \cap NF = K$ then $K$ is the midpoint of $BC$. But I still don't know how to prove that the theorem is true.

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  • $\begingroup$ Are you sure the condition is $\angle ACF=\angle ABC$? Because, I would expect (by symmetry with the other condition $\angle AEB=\angle ACB$) that the condition is $\angle AFC=\angle ABC$. $\endgroup$ – Julian Mejia May 29 at 3:08
  • $\begingroup$ I fixed the problem. $\endgroup$ – Lê Thành Đạt May 29 at 3:21
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Let $NF$ intersect $BC,AC$ at $K,L$ respectively. $EF \cap BC = D$. We will show that $K$ is the midpoint of $BC$. By Menalaus theorem, $$\dfrac{AN}{BN}\cdot \dfrac{BK}{CK}\cdot\dfrac{CL}{AL}=1.$$

Therefore, it suffices to show $$\dfrac{AN}{BN}\cdot\dfrac{CL}{AL}=1.$$

Now, $\dfrac{AN}{BN}=\dfrac{CD}{BC}$ and $\dfrac{CL}{AL}=\dfrac{CN}{AF}=\dfrac{\frac{AD\cdot BC}{BD}}{AF}=\dfrac{AD\cdot BC}{AF\cdot BD}$ since $AD$ and $CN$ are parallel.

Thus it suffices to show $$\dfrac{CD\cdot AD}{AF\cdot BD}=1.$$

But it is clear that $\triangle ABD\sim \triangle ACF \Longrightarrow \dfrac{AD}{AF}=\dfrac{AB}{AC}=\dfrac{BD}{CD},$ the latter being true due to the Angle Bisector Theorem.

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  • $\begingroup$ What is point $D$? $\endgroup$ – Lê Thành Đạt May 29 at 15:38

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