Answer :

Need to prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals

ABCD is a rhombus in which diagonals AC and BD intersect at point O.

We need to prove AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + DB^{2}

⇒ In Δ AOB; AB^{2} = AO^{2} + BO^{2}

⇒ In Δ BOC; BC^{2} = CO^{2} + BO^{2}

⇒ In Δ COD; CD^{2} = DO^{2} + CO^{2}

⇒ In Δ AOD; AD^{2} = DO^{2} + AO^{2}

⇒ Adding the above 4 equations we get

⇒ AB^{2} + BC^{2} + CD^{2} + DA^{2 =} AO^{2} + BO^{2} + CO^{2} + BO^{2} + DO^{2} + CO^{2} + DO^{2} + AO^{2}

⇒ = 2(AO^{2} + BO^{2} + CO^{2} + DO^{2})

Since, AO^{2} = CO^{2} and BO^{2} = DO^{2}

= 2(2 AO^{2} + 2 BO^{2})

= 4(AO^{2} + BO^{2}) ……eq(1)

Now, let us take the sum of squares of diagonals

⇒ AC^{2} + DB^{2} = (AO + CO)^{2} + (DO+ BO)^{2}

= (2AO)^{2} + (2DO)^{2}

= 4 AO^{2} + 4 BO^{2} ……eq(2)

From eq(1) and eq(2) we get

⇒ AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + DB^{2}

Hence, proved

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