How do you prove that sum of the squares on the sides of a rhombus is equal to the sum of squares on its diagonals?

1 Answer
CW
Jan 28, 2017

see explanation.

Explanation:

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Some of the properties of a rhombus :
#a)# all sides have equal length :
#=> AB=BC=CD=DA#
#b)# the two diagonals bisect each other at right angle:
#=> OA=OC, => AC=2OA=2OC#
Similarly, #OB=OD, => BD=2OB=2OD#

In #DeltaAOB#, as #angleAOB=90^@#,
by Pythagorean theorem, we know that #AB^2=OA^2+OB^2#
Let the sum of squares of the sides be #S_s#,
#=> S_s=4AB^2#

Let the sum of squares of the diagonals be #S_d#,
#S_d# is given by :
#S_d=AC^2+BD^2=(2OA)^2+(2OB)^2#
#=> S_d=4OA^2+4OB^2=4(OA^2+OB^2)=4AB^2#

#=> S_d=S_s# (proved)

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