MHB Where Do the Diagonals of a Parallelogram Intersect?

evinda
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Hello! (Wave)

Suppose that a parallelogram $ABCD$ has vertices $A=(0,0)$ and $B=(1,0)$. In terms of $C=(x,y)$, find the position of $D$ and where the two diagonals will intersect.

Then we will have something like that:

View attachment 5378How can we find the coordinates of $M$, i.e. the point at which the two diagonals intersect?
 

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Note that $\overrightarrow{AC}+\overrightarrow{AB}=\overrightarrow{AD}=2\overrightarrow{AM}$. You should use the connection between the coordinates of points and the coordinates of vectors, as well as rules for operations on vectors in terms of coordinates.
 
Does it hold that $\vec{AD}=2\vec{AM}$ from the fact that each diagonal of a parallelogram separates it into two equal triangles?I found that $D=(x \pm 1, y)$.

Let $M=(\mu_1, \mu_2)$.

So we have $\vec{AD}=(x \pm 1, y)$ and $\vec{AM}=(\mu_1, \mu_2)$.

So $\vec{AD}=2\vec{AM} \Rightarrow x \pm 1= 2 \mu_1 \text{ and } y=2 \mu_2 \Rightarrow \mu_1=\frac{x \pm 1}{2} \text{ and } \mu_2=\frac{y}{2}$

Thus $M=\left( \frac{x \pm 1}{2}, \frac{y}{2} \right)$. Right?
 
evinda said:
Does it hold that $\vec{AD}=2\vec{AM}$ from the fact that each diagonal of a parallelogram separates it into two equal triangles?
Yes, but it's easier to refer to the known fact that diagonals in a parallelogram are divided in half by the intersection point.

evinda said:
I found that $D=(x \pm 1, y)$.
It should be $D=(x + 1, y)$ because the $x$-coordinate of $B$ is $1$, not $-1$. The rest is correct.
 
Evgeny.Makarov said:
Yes, but it's easier to refer to the known fact that diagonals in a parallelogram are divided in half by the intersection point.
Is this known as a theorem?

Evgeny.Makarov said:
It should be $D=(x + 1, y)$ because the $x$-coordinate of $B$ is $1$, not $-1$. The rest is correct.

Couldn't we have D=(x-1,y) if we put D at the position I put C at the graph that I have drawn and C at the position of D?
Or am I wrong?
 
evinda said:
Is this known as a theorem?
Yes.

evinda said:
Couldn't we have D=(x-1,y) if we put D at the position I put C at the graph that I have drawn and C at the position of D?
You are right, it depends on the order of vertices. In fact, calling a quadrangle (and, more generally, a polygon) $XYZW$ usually means that the order of vertices is $X,Y,Z,W$ if they are visited in the counterclockwise or clockwise order. Using this rule, the parallelogram in post #1 should be called $ABDC$.
 
Evgeny.Makarov said:
Yes.

Ah, I see... (Nod)

Evgeny.Makarov said:
You are right, it depends on the order of vertices. In fact, calling a quadrangle (and, more generally, a polygon) $XYZW$ usually means that the order of vertices is $X,Y,Z,W$ if they are visited in the counterclockwise or clockwise order. Using this rule, the parallelogram in post #1 should be called $ABDC$.

So $X$ has to be connected with $Y$, $Y$ with $Z$, $Z$ with $W$ and $W$ with $X$, right?

I considered the parallelogram ABCD and then I got that $\vec{AC}=2 \vec{AM} \Rightarrow (x,y)=(2 \mu_1, 2 \mu_2)
\Rightarrow \mu_1=\frac{x}{2}$ and $\mu_2=\frac{y}{2}$.

So $M=\left( \frac{x}{2}, \frac{y}{2} \right)$, right?
 
evinda said:
So $X$ has to be connected with $Y$, $Y$ with $Z$, $Z$ with $W$ and $W$ with $X$, right?
That's how it is usually done.

evinda said:
I considered the parallelogram ABCD and then I got that $\vec{AC}=2 \vec{AM} \Rightarrow (x,y)=(2 \mu_1, 2 \mu_2)
\Rightarrow \mu_1=\frac{x}{2}$ and $\mu_2=\frac{y}{2}$.

So $M=\left( \frac{x}{2}, \frac{y}{2} \right)$, right?
Correct.
 
Great... Thanks a lot! (Smirk)
 
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