Question

ABCD is a rhombus. prove that 4AB^2=BC^2+BD^2

Answer 1

Let ABCD a rhombus..

AC and BD are diagonals..

We know that,

AB=BC=CD=DA

If O is at centre,

AO= OC= AC/2

BO= OD = BD/2

In Triangle AOB,

=>AB^2 = AO^2 +BO^2

=>AB^2 = (AC/2)^2 + (BD/2)^2

=>AB^2 = AC^2/4 + BD^2/4

=>AB^2 = AC^2+BD^2/4

=> 4AB^2 = AC^2 + BD^2

HENCE PROVED

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