# Vector proof for midpoints of 2 sides and diagonal intersection

Point $A, B, C, D$ have position vectors $a, b, c, d$ respectively relative to an origin O.

If $P$ divides $AB$ in the ratio $1:2$ and $Q$ divides $CD$ in the ratio $1:2$, obtain an expression for the position vector of $X$, where $X$ is the midpoint of $PQ$.

If $ABCD$ is a parallelogram show that $X$ is the point in which the diagonals $AC$ and $BD$ intersect.

I did the following diagram to work out this problem.

I figured that to get $\overrightarrow{OX}$, I need $\overrightarrow{PQ}$, $\overrightarrow{PX}$ and $\overrightarrow{AP}$, which I got as below,

\begin{align} \overrightarrow{PQ} &= \overrightarrow{PA} + \overrightarrow{AD} + \overrightarrow{DQ} \\ &= -\frac{2}{3} a - \frac{1}{3} b + \frac{2}{3} c + \frac{1}{3} d \\ \\ \overrightarrow{PX} &= \frac{1}{2}\overrightarrow{PQ} \\ &= -\frac{1}{3} a - \frac{1}{6} b + \frac{1}{3} c + \frac{1}{6} d \\ \\ \overrightarrow{AP} + \overrightarrow{PX} &= \overrightarrow{AX} \\ \overrightarrow{AX} &= \frac{1}{3} \overrightarrow{AB} + \overrightarrow{PX} \\ &= -\frac{2}{3}a + \frac{1}{6}b + \frac{1}{3}c + \frac{1}{6}d \\ \\ \overrightarrow{OX} &= \overrightarrow{OA} + \overrightarrow{AX} \\ &= \frac{1}{3}a + \frac{1}{6}b + \frac{1}{3}c + \frac{1}{6}d \\ &= \frac16(2a+b+2c+d) \end{align}

For the second part I figure I need to show that $X$ lies on $AC$ and $BD$. ie:- $\overrightarrow{AX} = k\overrightarrow{AC} |k\overrightarrow{XC}$ and $\overrightarrow{BX} = k\overrightarrow{BC} |k\overrightarrow{XD}$.

I tried doing this but end up with vectors without all the components. So I am unable to represent the vectors as multiples of each other. What am I missing? How do you solve the last part of this question?

In a parallelogram, you have $b-a=c-d$, so $b-a+d-c=0$. Adding one twelfth of that to your last displayed equation yields $\overrightarrow{OX}=\frac14(a+b+c+d)$, which is the midpoint of the parallelogram and the point where the diagonals intersect.
Thanks. Sorry if my question sounds stupid! Can you please clarify how you arrived at one twelfth? The equation I got is $\overrightarrow{OX} = \frac14(2a+b+2c+d)$, As per your suggestion, if $\overrightarrow{OX}=\frac14(a+b+c+d)$, Shouldn't $\overrightarrow{OX}$ be unique? – mathguy80 Jul 22 '11 at 12:46
I don't understand. You added the line $=\frac16(2a+b+2c+d)$ in an edit after I wrote my answer; this is correct. I don't know how you arrived at $\overrightarrow{OX} = \frac14(2a+b+2c+d)$. This is $\frac32\overrightarrow{OX}$, as you can see from $\overrightarrow{OX}=\frac16(2a+b+2c+d)$. If you add one twelfth of $0=b-a+d-c$ to $\overrightarrow{OX}=\frac16(2a+b+2c+d)$, the result is $\overrightarrow{OX}=\frac14(a+b+c+d)$, since $\frac26-\frac1{12}=\frac16+\frac1{12}=\frac14$. – joriki Jul 22 '11 at 13:05
I see what you mean about the one twelfths now. I had interpreted it as add one twelfth + $\overrightarrow{OX}$ instead of add both the equations together which makes more sense now. – mathguy80 Jul 22 '11 at 13:15