# Why is this the condition of collinearity?

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

Let $ABC$ be a triangle and let $A'\in BC$, $B'\in AC$ and $C'\in AB$. I want to show the following:

If $A',B',C'\neq A,B,C$ and either zero or two of the points $A',B',C'$ are at one side of triangle, then $A',B',C'$ are collinear iff $\frac{|AC'|}{|C'B|}=\frac{|BA'|}{|A'C|}=\frac{|CB'|}{|B'A|}=1$.

Could you give me a hint how we could show that?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!!

Let $ABC$ be a triangle and let $A'\in BC$, $B'\in AC$ and $C'\in AB$. I want to show the following:

If $A',B',C'\neq A,B,C$ and either zero or two of the points $A',B',C'$ are at one side of triangle, then $A',B',C'$ are collinear iff $\frac{|AC'|}{|C'B|}=\frac{|BA'|}{|A'C|}=\frac{|CB'|}{|B'A|}=1$.

Could you give me a hint how we could show that?
Hey mathmari!!

Three points on the respective sides of a triangle won't be on the same line.
Can you perhaps clarify what is intended?

#### mathmari

##### Well-known member
MHB Site Helper
Hey mathmari!!

Three points on the respective sides of a triangle won't be on the same line.
Can you perhaps clarify what is intended?
The original statement is the following:

Maybe the first sentence is not related to the second question?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
The original statement is the following:

Maybe the first sentence is not related to the second question?
As an observation, it looks like a product was intended instead of a set of equalities.

Anyway, let's pick an example.
Whether the first sentence applies or not, the following is an example that satisfies the first sentence and the conditions of question (b).
\begin{tikzpicture}[blue]
\coordinate (A) at (0,0);
\coordinate (C') at (3,0);
\coordinate (B) at (6,0);
\coordinate (A') at (5,1.5);
\coordinate (C) at (4,3);
\coordinate (B') at (2,1.5);
\draw[blue, ultra thick] (A) -- (B) -- (C) -- cycle;
\path (A) node[below left] {A} -- (B) node[below right] {B} -- (C) node[above] {C};
\draw[green!70!black, thick] (A') node[above right] {A'} -- (B') node[above left] {B'} -- (C') node[below] {C'} -- cycle;
\end{tikzpicture}
I've picked each of A', B', and C' to be exactly halfway, so that all the fractions are 1, and so that their product is also 1.

It doesn't look as if A', B', and C' are on the same line are they?
And that's what colinear ('kollinear' in German) means, doesn't it?

#### mathmari

##### Well-known member
MHB Site Helper
I have an idea...

Using the definitions of the first question we have the following:

$$|AC'|=|C'-A|=|A+\lambda (B-A)-A|=|\lambda (B-A)|=\lambda |B-A|$$
$$|C'B|=|B-C'|=|B-(A+\lambda (B-A))|=|B-A-\lambda (B-A)|=|(B-A)-\lambda (B-A)|=(1-\lambda )|B-A|$$
$$|BA'|=|A'-B|=|B+\mu (C-B)-B|=|\mu (C-B)|=\mu |C-B|$$
$$|A'C|=|C-A'|=|C-(B+\mu (C-B))|=|C-B-\mu (C-B)|=|(C-B)-\mu (C-B)|=(1-\mu )|C-B|$$
$$|CB'|=|B'-C|=|C+\nu (A-C)-C|=|\nu (A-C)|=\nu |A-C|$$
$$|B'A|=|B'-A|=|C+\nu (A-C)-A|=|-(A-C)+\nu (A-C)|=|-(A-C)+\nu (A-C)|=-(1-\nu )|A-C|$$

Replacing this we get the following:

$$\frac{\lambda |B-A|}{(1-\lambda )|B-A|}\cdot \frac{\mu |C-B|}{(1-\mu )|C-B|}\cdot \frac{\nu |A-C|}{-(1-\nu )|A-C|}=\frac{\lambda}{1-\lambda }\cdot \frac{\mu }{1-\mu }\cdot \frac{\nu }{-(1-\nu)}=-\frac{\lambda\cdot \mu \cdot \nu}{(1-\lambda)\cdot (1-\mu )\cdot (1-\nu) }$$

From the first question we have that $A',B',C'$ are collinear iff $\lambda \mu \nu +(1-\lambda )(1-\mu )(1-\nu)=0 \Rightarrow \lambda \mu \nu =-(1-\lambda )(1-\mu )(1-\nu) \Rightarrow -\frac{\lambda \mu \nu }{(1-\lambda )(1-\mu )(1-\nu)}=1$.

So we get the desired result, right?

#### Opalg

##### MHB Oldtimer
Staff member
I have an idea...

Using the definitions of the first question we have the following:

$$|AC'|=|C'-A|=|A+\lambda (B-A)-A|=|\lambda (B-A)|=\lambda |B-A|$$
$$|C'B|=|B-C'|=|B-(A+\lambda (B-A))|=|B-A-\lambda (B-A)|=|(B-A)-\lambda (B-A)|=(1-\lambda )|B-A|$$
$$|BA'|=|A'-B|=|B+\mu (C-B)-B|=|\mu (C-B)|=\mu |C-B|$$
$$|A'C|=|C-A'|=|C-(B+\mu (C-B))|=|C-B-\mu (C-B)|=|(C-B)-\mu (C-B)|=(1-\mu )|C-B|$$
$$|CB'|=|B'-C|=|C+\nu (A-C)-C|=|\nu (A-C)|=\nu |A-C|$$
$$|B'A|=|B'-A|=|C+\nu (A-C)-A|=|-(A-C)+\nu (A-C)|=|-(A-C)+\nu (A-C)|=-(1-\nu )|A-C|$$

Replacing this we get the following:

$$\frac{\lambda |B-A|}{(1-\lambda )|B-A|}\cdot \frac{\mu |C-B|}{(1-\mu )|C-B|}\cdot \frac{\nu |A-C|}{-(1-\nu )|A-C|}=\frac{\lambda}{1-\lambda }\cdot \frac{\mu }{1-\mu }\cdot \frac{\nu }{-(1-\nu)}=-\frac{\lambda\cdot \mu \cdot \nu}{(1-\lambda)\cdot (1-\mu )\cdot (1-\nu) }$$

From the first question we have that $A',B',C'$ are collinear iff $\lambda \mu \nu +(1-\lambda )(1-\mu )(1-\nu)=0 \Rightarrow \lambda \mu \nu =-(1-\lambda )(1-\mu )(1-\nu) \Rightarrow -\frac{\lambda \mu \nu }{(1-\lambda )(1-\mu )(1-\nu)}=1$.

So we get the desired result, right?
Yes, that is correct. Notice that the question says that either none or exactly two of the points $A'$, $B'$, $C'$ lie on a side of the triangle. This implies that either one or all three of these points lie on an extension of a side of the triangle.

The other important point here is that notations such as $|AC'|$ are used in this question to denote signed distances. Thus $|AC'| = -|C'A|$. That is why I like Serena 's construction in comment #4 does not provide a counterexample to this result.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes, that is correct. Notice that the question says that either none or exactly two of the points $A'$, $B'$, $C'$ lie on a side of the triangle. This implies that either one or all three of these points lie on an extension of a side of the triangle.

The other important point here is that notations such as $|AC'|$ are used in this question to denote signed distances. Thus $|AC'| = -|C'A|$. That is why I like Serena 's construction in comment #4 does not provide a counterexample to this result.
Aha! So we have something like:
\begin{tikzpicture}
\def\bx{6};
\def\cx{4};
\def\cy{3};
\def\lambda{(1/2)};
\def\mu{(3/4)};
\def\nu{-(1/2)};
\coordinate (A) at (0,0);
\coordinate (C') at ({\lambda*\bx},0);
\coordinate (B) at (\bx,0);
\coordinate (A') at ({\mu*\cx+(1-\mu)*\bx},{\mu*\cy});
\coordinate (C) at (\cx,\cy);
\coordinate (B') at ({(1-\nu)*\cx},{(1-\nu)*\cy});
\draw[blue, ultra thick] (A) node[below left] {A} -- (B) node[below right] {B} -- (C) node[above left] {C} -- cycle;
\draw[green!70!black, thick] (A') node
{A'} -- (B') node[above left] {B'} -- (C') node[below] {C'} -- cycle;
\draw[green!70!black, thick, dashed] (C) -- (B');
\end{tikzpicture}​