# A geometry problem - prove that 3 points are collinear.

Let $ABCD$ be a parallelogram and $G$ the center of gravity(the intersection point of the medians) for the $\triangle ABC$. $M \in AD$ and $D \in NC$.

Prove that $G,M,N$ are collinear if and only if $\displaystyle \frac{CN}{ND}-\frac{AM}{MD}=\frac{1}{2}.$

I don't know anything about this problem, it seems to hard for me. How can I use the notion of collinearity knowing that raport ?

Thanks!

• Do you mean "$N \in DC$" (instead of "$D\in NC$")? If not, then where is $N$?
– Blue
Commented Jan 11, 2013 at 13:43
• Point $N$ is used before being defined. Statement $D \in NC$ needs correction. Commented Jan 11, 2013 at 14:42
• If $N$ is on $CD$ extended beyond the parallelogram then $G, M, N$ can be collinear and $D \in NC$. Commented Jan 11, 2013 at 14:45
• Well @Freddy, "everything's fine" now that you gave your personal interpretation and explanation, *not before" when we had to guess what/where is N...The OP is supposed to pose clear questions, it's not other members' task to guess. Commented Jan 11, 2013 at 20:53
• No @Iule, it's not OK as you wrote it at the beginning. NOW, after you added a diagram, it is ok and clear. Commented Jan 11, 2013 at 20:57

The setup of the problem is "affine invariant", which implies that we may assume $ABCD$ to be a square in the $(x,y)$-plane. In other words, we may introduce coordinates $(x,y)$ such that $$A=(0,0), \ B=(1,0), \ C=(1,1), \ D=(0,1), \ N=(u,1), \ M=(0,v)\ ,$$ where the numbers $u$, $v\in{\mathbb R}$ have certain values. It follows that $G=\bigl({2\over3},{1\over3}\bigr)$ and therefore $$\vec{GM}=\left(-{2\over3}, v-{1\over3}\right),\quad \vec{GN}=\left(u-{2\over3}, {2\over3}\right)\ .$$ The three points $G$, $M$, $N$ are in line iff the "vector product" $\vec{GM}\wedge\vec{GN}=0$, which means that $u$ and $v$ have to satisfy $$-uv +{2\over3} v+{1\over3}u-{2\over3}=0\ .\qquad(*)$$ On the other hand $${CN\over ND}={u-1\over -u}\ ,\quad {AM\over MD}={v\over 1-v}\ ,$$ and it is easily verified that ${CN\over ND}-{AM\over MD}={1\over2}$ is equivalent with $(*)$.

Let the line through $G$ parallel to $AB$ intersect $BC$ at $X$, and the line through $G$ parallel to $BC$ intersect $AB$ at $Y$. Mark the length of $GX$ as $a$, and the length of $GY$ as $b$. Its easy to prove that $a = AB/3$ and $b = BC/3$ by solving the system of equations of lines AG and CG.

Let $T$ be the intersection of $GY$ and $CD$. By using the similar triangles $NDM$ and $NTG$ we get:

$\frac{ND}{MD} =\frac{NT}{GT} =\frac{ND + 2a}{2b} = \frac{ND}{2b} + \frac{a}{b}$

multiply by: $\frac{3b}{ND}$

$\frac{3b}{MD} = \frac{3}{2} + \frac{3a}{ND}$ (1)

Rearranging the original equation and using (1) we get:

$\frac{CN}{ND} - \frac{AM}{MD} = \frac{3a + ND}{ND} - \frac{3b - MD}{MD} = \frac{3a}{ND} - \frac{3b}{MD} + 2 = \frac{1}{2}$

This was the 'only if' part of the 'if and only if'. For the 'if' part, because we only used equalities, and every equation was equivalent to the previous, we can say that from

$\frac{CN}{ND} - \frac{AM}{MD} = \frac{1}{2}$

follows that

$\frac{ND}{MD} =\frac{NT}{GT}$

This, along with the fact that the corresponding angles are equal due to parallel lines, shows that triangles $NDM$ and $NTG$ are similar.

From the similarity, we get that the angle $\angle MND = \angle GNT$.

$M$, $D$, and $T$ are collinear, so $\angle GND = \angle GNT$.

From the last two sentences: $\angle MND = \angle GND$, which is a proof that $N$, $M$, and $G$ are collinear.

• If the intersection of $GY$ and $CD$ is $T$, $NDM$ is similar to $NTG$ Commented Jan 11, 2013 at 21:10
• I didn't write more specifically because that point had no name. Commented Jan 11, 2013 at 21:11
• Wait a second, I'll edit my answer Commented Jan 11, 2013 at 21:35
• This answer looks essentially correct to me. I don't understand the two downvotes. If you ask me, it's the question that is badly worded: it should be "Show that if $M$ is the intersection of $AD$ and $GN$, then $CN/ND - AM/MD = 1/2$". The key observation is that $G$ lies on $DB$, with $DG = 2GB$. Commented Jan 11, 2013 at 21:58