Can somebody give a solution just with euclidean geometry to this problem?
Here's the problem:
If $A$, $C$, $E$ are collinear and $B$, $D$, $F$ are collinear (as you see in the picture) and if $$\frac{CE}{AC}=\frac{DF}{BD}=\lambda$$ prove that $M$, $N$, $P$ are collinear, where $M$, $N$, $P$ are the midpoints of the sides $AB$, $CD$, $EF$, respectively. (The picture shows $\lambda=7/3$.)
A solution:
Consider $M=(0,0)$, $A=(-a,0)$, $B=(a,0)$. Let $C=(b,c)$ and $D=(e,g)$; then $N=(\frac12(b+e),\frac12(c+g))$. Because $\overrightarrow{CE}=\lambda\overrightarrow{CA}$, we have $E=((\lambda+1)b+\lambda a,(\lambda+1)c)$. Similarly, because $\overrightarrow{DF}=\lambda \overrightarrow{BD}$, we have $F=((\lambda+1)e-\lambda a,(\lambda+1)g)$. Therefore, $P = ((\lambda+1)(b+e),(\lambda+1)(c+g))$, so $\overrightarrow{MP}=(\lambda+1)\overrightarrow{MN}$, showing $M$, $N$, $P$ collinear. $\square$
A solution only for $a=b$
Line segment $AB$ and points $D$ $F$ on the same side of $AB$ such that $AD=BF$ The extensions of $AD$ and $BF$ meet at a point $C$. Draw the circle the goes through the points $A,B,C$(Let's call it $Q$) .Let $M,N$ as the midpoints of the $AB,DF$ respectively. $Lemma$:the perpendicular bisectors of AB and DF respectively meet at a point $P$, $P\in(Q)$:Proof $($ if the perpendicular bisector of $AB$ meets $Q$ at $P$ , the triangles $ADP$ and $BFP$ are equal to each other, $PD=PF$ so $PN$ is the perpendicular bisector of DF $)$ Let $PY,PT$ perpendicular to the lines $AC,BC$, $T\in(lineAC)$ $Y\in(lineBC)$. According to Simpsons theorem $M,N,Y$ are collinear. We will show that $N\in(MY)\iff{N\in(NY)}\iff{DFCP:inscribable}\iff{\angle{ACB}=\angle{DPF}}\iff{\angle{ACB}=\angle{WPV}}\iff{\stackrel\frown{AW}+\stackrel\frown{WB}=\stackrel\frown{WB}+\stackrel\frown{BV}}\iff{\stackrel\frown{AW}}=\stackrel\frown{BV}\iff{\angle{APD}=\angle{BPF}}$ which is true, because the are same angles of the equal triangles $ADP$ and $BPF$ (Note:$W=line(PD)\cap{Q}$ ,$V=line(PF)\cap{Q}$. Similarly we could prove that every point with similar properties to those, which described N, belong as well to MY, therefore are collinear. $\square$
Let $A = (0,0), C = (1, 0), E = (1 + \lambda, 0)$. Let $B, D, F$ lie on the line $y = mx + c$ s.t. $B$ lies on $y-$axis. Hence, $B = (0,c)$. Let $F = (x_0, mx_0 + c)$. Therefore, $D$ divides the line $BF$ in the ratio $1 : \lambda$ internally. Hence, $D = (\frac{x_0}{1 + \lambda}, \frac{\lambda c + mx_0 + c}{1 + \lambda})$.
Now, $M = \left(0 , \frac c2 \right)$, $N = \left(\frac{1 + \lambda + x_0}{2(1 + \lambda)}, \frac{mx_0 + c(1 + \lambda)}{2(1 + \lambda)}\right) $, $P = \left( \frac{1 + \lambda + x_0}{2}, \frac{mx_0 + c}{2} \right)$.
\begin{align} \text{Area of} \Delta MNP &= \det \begin{bmatrix}0 & \frac{c}{2} & 1 \\ \frac{1 + \lambda +x_0}{2(1 + \lambda)} & \frac{mx_0 + c(1 + \lambda)}{2(1 + \lambda)} & 1 \\ \frac{1 + \lambda + x_0}{2} & \frac{mx_0 + c}{2} & 1\end{bmatrix} \\ & = 0 \end{align} Hence, $M, N, P $ are collinear.