Need Assistance with algebra

[Paul]

My first question is, is my solution correct?

prove that the sum of the squares of the diagonals of any parallelogram is equal to the sum of the squares of its sides.

using vectors
parallelogram ABCD, where vectors AB= DC and AD=BC, with diagonals AC and BD

code:

here's what I did: let AD = a, AB = b, AC=c and BD=d so to clarify: a = side 1 b = side 2 (not parallel to side 1) c = diagonal 1 d = diagonal 2 c = a + b d = b - a (keep in mind they're all vectors) sum of the diagonals:     (a+b)^2 + (b-a)^2 = a^2 + b^2 + 2ab + b^2 + a^2 -2ab =2a^2 + 2b^2 since side 1 (a) = side 3 and side 2(b) = side 4 you can say that Side1^2 + side2^2 + side3^2 +side4^2 = diagonal1^2 + diagonal2^2 QED

I'm just not sure about my solution, as it deals with multiplication of vectors, and I don't know anything about the validity of that.

My second question is:

Quote:

If a line through the centre of a circle is perpendicular to a chord, prove that it intersects the chord at its midpoint

And I have no idea how to do that, first I thought dot product, but that didn't work out.

[Brightguy]

First of all, you want to be dealing with the magnitudes of the vectors (i.e., their length). The magnitude of vector a is denoted |a|.

Even still, your proof is not completely valid, since in general
|a+b|² ≠ |a|² + 2|a||b| + |b|²
But your proof is not that far off... here's a hint: remember the cosine law.

And as for geometric proofs, I found that often there was some construction that could be made to help with the proof. In this case, try drawing lines from the centre of the circle to the ends of the chord, and noticing that you have an isosceles triangle.