Question

15. ABCD is a rhombus and its two diagonals meet at
O, show that
ADP + DC+ CB2 + BA? = 4 DO2 + 4 AO2​

Answer 1

Consider the given rhombus ABCD

Let AC&BD be the diagonals of rhombus ABCD

In ΔAOD by phythagoras theorem,

AB² = AO² + DO²

By multiplying by 4 on both side and we get,

4AB² = 4AO² + 4DO²

We know that, the all sides of rhombus are equal

Then, AB=BC=CD=DA

So, AB² + BC² + CD² + DA²= 4AO² + 4DO²

Hence proved.

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