Q: Prove that each diagonal of a parallelogram bisects each other
How do I attempt this?
Can I find the midpoints of the diagonals, then if they're the same, get the distance between this midpoint and the vertices? If they're the same, have I proved it?
Thanks.
It can be proved using vectors. Consider a parallelogram ABCD with a,b,c,d as the position vectors of points A,B,C,D respectively.Quote:
Originally Posted by Glitch
so b-a=c-d
Rearrange it to get a+c=b+d
Divide both sides by 2, (a+c)/2=(b+d)/2
We see that the midpoints of AC and BD are identical.
...
I don't think I'm supposed to solve this with vectors, since I haven't learnt about them yet.
use congruent triangles, then.
There's the problem- if you don't show any work at all, we have no idea what you can use!Quote:
Originally Posted by Glitch
In addition to "congruent triangles" (basic geometry) and "vectors", I might use "coordinate geometry". We can always set up a coordinate system so that one vertex is at (0, 0), another at (a, 0), a third at (b, c) and then the fourth must be at (a+ b, c). Now, it is easy to find the midpoint of each diagonal- that is, if you can use coordinate geometry- and we don't know that.
I just wanted to know if my reasoning was sound.Quote:
Originally Posted by HallsofIvy