Diagram shows quadrilateral ABCD, with OA = (-6,3) , OB = (5,5), OC = (7,-2) , OD = (-4,-6). Show that midpoints P, Q, R & S form a parallelogram.

Question

The diagram below shows the quadrilateral ABCD, with OA = (-6,3) , OB = (5,5), OC = (7,-2) , OD = (-4,-6).

Show that the midpoints P, Q, R & S form a parallelogram.

My working out so far:

To show that PQRS forms a parallelogram, we must show:

PQ//SR and SP//RQ

I'm unsure how to find each of the position vectors of P, Q, R and S. For example to find OP would you first find AB?

^^ Given this is the case, I have taken

AB = OB + AO

= OB - OA

= (5i + 5j) - (-6i + 3j) = -i +2j

I don't know how to continue the proof from here. How would you explain the steps of how to prove this? This is from a Grade 11 exam paper.

Answer 1

First of all, I recommend that when you can, without too much effort, solve a general problem (for which the question pertains to a specific case), you should always do so.

Here, the general case may actually be simpler!

Let (for instance), $\vec{OA} = \mathbf a, \vec{OP} = \mathbf p$ and so forth. These are position vectors relative to the origin.

First find the position vectors for $\mathbf{p,q,r,s}$ respectively as $\frac 12\mathbf{(a+b)},\frac 12\mathbf{(b+c)}, \frac 12\mathbf{(c+d)}, \frac 12\mathbf{(d+a)}$

Next find the vectors for $\vec{PQ}, \vec{QR},\vec{RS},\vec{SP}$ respectively, for example:

$\vec{PQ} = \vec{OQ} - \vec{OP} = \frac 12\mathbf{(b+c)} - \frac 12\mathbf{(a+b)} = \frac 12\mathbf{(c-a)}$.

Similarly, $\vec{QR} = \frac 12\mathbf{(d-b)}, \vec{RS} = \frac 12\mathbf{(a-c)}, \vec{SP} = \frac 12\mathbf{(b-d)}$. You shouldn't have to work each of these out, just use the cyclical symmetry of the vertices!

Finally, note that $\vec{PQ} = -\vec{RS}$ and $\vec{QR} = -\vec{SP}$, which yields: $|PQ| = |RS|, PQ \parallel RS$ and $|QR| = |SP|, QR \parallel SP$ which is sufficient to establish that $PQRS$ is a parallelogram. (To be precise, even one set of line segments of equal length and parallel would do it, but no harm in being thorough).

Note that the actual numerical coordinates of the points in the specific question never needed to be used because it's just a case of a more general theorem called Varignon's Theorem. The proof I gave above establishes the general theorem.

Answer 2

We have

$$\vec{AP} = \frac12 \vec{AB}$$

$$\vec{OP}-\vec{OA}=\frac12 \left(\vec{OB} - \vec{OA} \right)$$

$$\vec{OP}=\frac12 \left( \vec{OB}+\vec{OA} \right)$$

Similarly for the other midpoints. That is if you wish, you can evaluate the midpoints.

We have \begin{align}\vec{PQ}&=\vec{OQ}-\vec{OP}\\&=\frac12(\vec{OC}+\vec{OB}) - \frac12(\vec{OA}+\vec{OB})\\ &=\frac12 \left( \vec{OC}-\vec{OA}\right)\\ &=\frac12(\vec{OC}+\vec{OD}) - \frac12(\vec{OA}+\vec{OD})\\ &= \vec{OR}-\vec{OS}\\ &=\vec{SR}\end{align}

I will leave the rest to you.

Answer 3

Because $$SP=\frac{1}{2}BD=QR,$$ $$SP||BD$$ and $$QR||BD,$$ which gives $SP=QR$ and $SP||QR,$ which says that $PQRS$ is a parallelogram.