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Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Q7   Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the
       squares of its diagonals.

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In \triangle AOB, by Pythagoras theorem,

AB^2=AO^2+BO^2..................1

In \triangle BOC, by Pythagoras theorem,

BC^2=BO^2+CO^2..................2

In \triangle COD, by Pythagoras theorem,

CD^2=CO^2+DO^2..................3

In \triangle AOD, by Pythagoras theorem,

AD^2=AO^2+DO^2..................4

Adding equation 1,2,3,4,we get

AB^2+BC^2+CD^2+AD^2=AO^2+BO^2+BO^2+CO^2+CO^2+DO^2+AO^2+DO^2

AB^2+BC^2+CD^2+AD^2=2(AO^2+BO^2+CO^2+DO^2)

\Rightarrow AB^2+BC^2+CD^2+AD^2=2(2.AO^2+2.BO^2)        (AO=CO  and BO=DO)

\Rightarrow AB^2+BC^2+CD^2+AD^2=4(AO^2+BO^2)

\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC}{2})^2+(\frac{BD}{2})^2)

\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC^2}{4})+(\frac{BD^2}{4}))

\Rightarrow AB^2+BC^2+CD^2+AD^2=AC^2+BD^2

Hence proved

 

 

 

 

 

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