Dark Horse's Question Thread

[Dark Horse]

First Things first: I have a vector proof question which is annoying me:

Prove that if the midpoints of the sides of a quadrilateral are joined, then a parallelogram is formed.

If it said rectangle, it would be easy, but if it is just any quadilateral, then can you make the assumption that opposite sides are the same vecttor?

A worked solution or instruction on how to prove it would be awesome!

Hope I'm not asking for too much

[the.watchman]

Lol, I have a non-vector proof, do you want that?

[/0]

I'd love to see your proof the.watchman


Yeah, since we can't assume any of the sides of the quadrilateral are congruent, we could start by giving each of them their own vector.









Also notice that

Then





But since



So, since and are exactly equal, two sides of the EFGH are congruent, which implies that the other two are as well (by symmetry). So EFGH is a parallelogram.

[the.watchman]

In a quadrilateral ABCD (named clockwise from top left), let the midpoints of AB, BC, CD & DA be E, F, G, H respectively

Join AC and BD

In AES and ABD,

Because (given)

Therefore

Because HAE = DAB (common angle)

Therefore AEH is similar to ABD (PAP)

Because AHE = ADB (corresponding angles in similar triangles are equal)

Therefore EH is parallel to BD

REPEAT THE ABOVE STEPS FOR THE OTHER SIDES TO OBTAIN:

FG parallel to BD, HG parallel to AC and EF parallel to AC

From the above, EH is parallel to FG and HG is parallel to EF

So EFGH is a parallelogram

[/0]

Ah very nice. I think euclidean geometry requires more skill than vector geometry

[Dark Horse]

Wow, those are both great answers! Thanks /0 and the.watchman! +1 karma:)

[the.watchman]

Ah very nice. I think euclidean geometry requires more skill than vector geometry

Thanks, that was the question that got me a unimelb bhp maths comp prize