prove by vectors that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
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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
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
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