Prove that the sum of the squares of the diagonals of a parallelogram = the sum of the squares of its sides.?

Answer 1

Think of the four corners as vectors, 0, A, B, A + B.

The square of the length of a vector X is X dot X.

Your left hand side is: ((A - B) dot (A - B)) + ((A + B) dot (A + B))
Your right hand side is 2(A dot A + B dot B)

((A - B) dot (A - B)) + ((A + B) dot (A + B))
= (A dot A - A dot B - B dot A + B dot B) + (A dot A + A dot B + B dot A + B dot B)
= 2(A dot A + B dot B)

Dan

Answer 2

Let L be the long side of the parallelogram (PG). Let S be the short side of the PG. Let H be the height of the PG. Let d be the short PG diagonal and D be the long PG diagonal. So that d^2 = (L - x)^2 + H^2 and D^2 = (L+x)^2 + H^2, where L - x = X the side of the rectangle with dimensions H by X inside the PG and so that S^2 = x^2 + H^2.

Prove: d^2 + D^2 = L^2 + S^2

d^2 + D^2 = (L - x)^2 + H^2 + (L+x)^2 + H^2 = [L^2 - 2Lx + x^2 + H^2] + [L^2 + 2Lx + x^2 + H^2 = 2[L^2 + x^2 + H^2] and S^2 = x^2 + H^2; so that d^2 + D^2 = 2[L^2 + S^2]

Sorry, I'm getting that the sum of squares of the diagonals (d and D) = 2 X the sum of the squares of the sides (S and L). This seems a reasonalbe result because it's clear H^2 has to be in there twice since H is the short side for the two right triangles having d and D as diagonals.

Answer 3

Let the sides be a,b,c,d where a=c and b=d because its a parallelogram.
Let the diagonals be e and f.
Let the angles of the parallelogram be X and Y (2 of each)
By the law of cosines:
e^2 = a^2 + b^2 - 2ab*cosX
f^2 = a^2 + b^2 - 2ab*cosY
e^2 + f^2 = 2(a^2+b^2) - 2ab(cosX + cosY)

but 2X + 2Y = 360 in a parallelogram.
X+Y = 180
X = 180 - Y
cosX = -cosY
cosX + cosY = 0
so:
e^2 + f^2 = 2(a^2+b^2) = a^2+b^2+c^2+d^2

Answer 4

Sorry,I had earlier made a mess of it taking ABCD as a rectangle.Idid not know how it happened but am very sorry.Hope you wouldnot mind this lapse on my part.
Here is the correct proof
ABCD is a parallelogram with AC and BD as its diagonal.To prove that ,
AC^2+BD^2=AB^2+BC^2+CD^2+AD^2
Construction: PerpendicularsCE and DF are dropped from C and D on extended AB
Similarly perpendiculars AH and BG are dropped from A and B on extended CD and Cd respectively
Proof:In right-angled triangle ACE,
AC^2=AE^2+CE^2
=(AB+BE)^2+CE^2
=AB^2+BE^2+2AB.BE+BC^2-BE^2
=AB^2+BC^2+2AB.BE
again in right-angled triangle BFD,
BD^2=BF^2+DF^2
=(AB-AF)^2+AD^2-AF^2
=AB^2+AF^2-2AB.AF+aD^2-AF^2
=CD^2+2AB.BE+AD^2 [as AB=CD and AF=BE]
Therefore AC^2+BD^2
=AB^2+BC^2+2AB.BE+CD^2+AD^2-2AB.BE
=AB^2+BC^2+CD^2+AD^2 (Proved)