# Prove that the intersection of $ME$ and $NF$ is a fixed point.

$$\triangle ABC$$ 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.

• 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 $\angle AFC=\angle ABC$. – Julian Mejia May 29 '19 at 3:08
• I fixed the problem. – Lê Thành Đạt May 29 '19 at 3:21

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.
• What is point $D$? – Lê Thành Đạt May 29 '19 at 15:38