$ABCDEF :$ Regular hexagon and $M\in [AC]$ , $N\in [CE]$ such : $\frac{AM}{AC}=\frac{CN}{CE}=\lambda $ where $\lambda >0$.
Find $\lambda $ such $B,M$ and $N$ are colinear points?
I don't know how to start this problem.
Maybe we should use angle $120°$ and $60°$
I need ideas or hints to approach it! enter image description here
This solution by Apostolis Manoloudis :
But I don't understand why he starte with $AB=AM$ ? Can someone explain to me this solution If any one have another way drop here
(New redaction of this paragraph). You are right when doubting about the correctness of the reasoning of this person. No basis for saying that $AM=AP$ !!! Mr Manoloudis makes an erroneous circular reasoning (petitio principii). This could be caught up in the following way. Let us take the sides of the hexagon as unit length. First apply Al Kashi's theorem to triangle $ABC$ : $AC^2=AB^2+BC^2-2 AB.BC \cos(120°)$ giving $AC=\sqrt{3}$. Now, making the assumption that the reversed ratio $\tfrac{1}{\lambda}=\tfrac{AC}{AM}=\sqrt{3}$, then $AM=1$, therefore $AM=AB$, triangle $BAM$ is isosceles ; having summit angle $30°$, base angle $AMB$ is $75°$, and we can resume...
Remark : angle-chasing based proofs aren't appreciated by geometers because they are figure-dependent, thus prone to errors.
Here is a completely different proof using barycentrical coordinates in triangle $ACE$.
What are barycentrical coordinates ? They are based on the fact, proved in lectures, that every point $M$ in the plane is completely determined by the list of weights (= degrees of attraction) $(a,c,e)$ (normalized as percentages, i.e., with $a+c+e=1$) to be placed on $A$, $C$ and $E$ resp. in such a way that $M$ is the barycenter of these points with these weights, corresponding to notation
$$M=aA+cC+eE.$$
For example, the center of gravity of $ACE$ has baryc. coord. $(\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3})$, the midpoint of $AC$ has baryc. coord. $(\tfrac{1}{2},\tfrac{1}{2},0)$. Baryc. coordinates are all positive for points inside triangle $ACE$ whereas, for points outside this triangle, some coordinates are negative exhibiting a "repulsion" by some vertices.
As $M=(1-\lambda)A+\lambda C$, the barycentrical coord. of $M$ are
$$((1-\lambda), \ \lambda, \ 0)\tag{1}$$
For the same reason, relationship $N=(1-\lambda)C+\lambda E$ gives the following baryc. coord. for $N$ :
$$(0,\ (1-\lambda), \ \lambda)\tag{2}$$
Last point $B$ has baryc. coord. (see proof below) :
$$(2/3,\ 2/3, \ -1/3)\tag{3}$$
The necessary and sufficient condition for $M, \ N, \ B$ to be aligned is that the determinant of their baryc. coord. is zero:
$$\begin{vmatrix}0&(1-\lambda)&\lambda\\(1-\lambda)&\lambda&0\\2/3&2/3&-1/3\end{vmatrix}=0$$
Expanding, one gets $1-3\lambda^2=0$, therefore
($\lambda=- \tfrac{\sqrt{3}}{3}$ could be a solution if $M$, resp. $N$ weren't constrained to be in line segments $[AC]$, resp. $[CE]$).
Proof of relationship (3) :
$$2(\vec{BA}+\vec{BC})=\vec{BE} \ \iff \ 2(A-B+C-B)=E-B \ \iff \ B=\tfrac{2}{3}A+\tfrac{2}{3}C+(-\tfrac{1}{3})E$$