I have been coming up with problems for my year 12 coordinate geometry students. Using algebra to prove general statements about basic shapes like triangles and quadrilaterals.
I came up with this question: Given a quadrilateral with vertices:
$P:(0,0), Q:(a,b), R:(c,d), S: (e,f)$
Prove that if $PQ$ is parallel to $RS$, and $PR$ is parallel to $QS$ (the definition of a parallelogram), then PQ and RS have the same length.
This seems to be a basic quality of quadrilaterals, which if you made one edge horizontal you could easily solve. However, the equations you get from this simple looking problem are bizarre.
The two given relations (the lines being parallel) correspond to:
$(1): \frac{f-b}{e-a}=\frac{d}{c}$ and $(2): \frac{f-d}{e-c}=\frac{b}{a}$
And somehow we need to prove $(3): a^2+b^2=(e-c)^2+(f-d)^2$
If you multiply out the fractions, none of the terms correspond to anything in the final expanded equation we need to prove.
Is there some algebraic way to prove (3) from (1) and (2) without resorting to the properties of a parallelogram?