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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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