prove that sum of squares of the digonals of a parallelogram is eq. to the sum of the squares of its sides?

Answer 1

Look ma, no trig!

Consider parallelogram abcd where ab is the lower horizontal segment of length A, and segment ad of length B slants up and to the right. Let the (longer) diagonal (ac) be of length D and the other diagonal (bd) have length E. Finally, let the distance between the horizontal sides (ab, cd) be H (the altitude).

Now drop an altitude from the top left vertex (d) onto the bottom horizontal side (ab) at point X. Let the distance from point X to the bottom left vertex (a) be F. Note that if you drop an altitude from the top right vertex onto the bottom horizontal side (ab) at Y, that the distance from X to Y is A (this is just the projection of the top horizontal side (cd) onto the bottom horizontal side (ab)).

For Pete's sake, draw a picture!
And now we look at three right triangles:

H² + F² = B² (from ΔbYc or ΔaXd)

And D² = H² + (A+F)² (from ΔaYc)
And E² = H² + (A-F)² (from ΔdXb)
Add these last two to get:
D² + E² = 2H² + 2A² + 2F²

and now use the first identity to get:
D² + E² = 2A² + 2B²

QED

Answer 2

Picture a parellelogram with sides a and b and diagonals c1 and c2. The angles between a and b are t1 and t2. Recall that t1 and t2 will add up to 180 degrees. (The four angles in a parallelogram are t1, t2, t1 and t2 and the total degrees in a quadrilateral is 360 degrees, so t1 + t2=180.)

The law of cosines says that

c1^2 = a^2 + b^2 - 2ab cos t1 and
c2^2 = a^2 + b^2 - 2ab cos t2
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c1^2 + c2^2 = 2a^2 + 2b^2 - 2ab(cos t1 + cos t2)

but the cos t2 = cos (180-t1) = -cos t1, so the last term becomes zero, so

c1^2 + c2^2 = 2a^2 + 2b^2