Question

abcd is a rhombus.prove that ac^2+bd^2=4ab^2

Answer 1

Let ABCD is rhombus

AC and BD are diagonals

we know that,

AB=BC=CD=AD

If O is centre,

AO = OC = AC/2

BO = OD = BD/2

In triangle AOB,

AB² = AO² + BO²

=> AB² = (AC/2)² + (BD/2)²

=> AB² = AC²/4 + BD²/4

=> AB² = AC²+BD²/4

=> 4AB² = AC² + BD²

Hence proved

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Answer 2

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