# A geometric problem regarding collinearity

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. $$\square$$ • What have you tried? Can you attach the image? – Calvin Lin Apr 6 '20 at 14:54
• Now I think you can find the image. I do have a geometric solution which uses only euclidean geometry and a simple analytic one.The first one is very complicated so i was wondering if someone could present an easier solution to it. – John Kall Apr 6 '20 at 15:04
• A coordinate geometry approach works. $PN/NM = \lambda$. Can you take it from here? (If you're comfortable with vectors, that's much more obvious.) – Calvin Lin Apr 6 '20 at 15:09
• Can you add your solution? The coordinate geometry approach is very directly, literally write it out and everything falls into place. I've voted to reopen this question given that you're demonstrating your work. – Calvin Lin Apr 6 '20 at 15:09
• Which one should I demonstrate? The one with coordinates or the euclidean?Furthermore should I make a new post with them or demonstrate them in a comment like this one?(ApparentIy I am new here). – John Kall Apr 6 '20 at 15:39

## 1 Answer

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.