Question

The diagonal $AC$ and $BD$ of a rhombus intersect each other at $O$. Prove that: $A{B}^2+B{C}^2+C{D}^2+D{A}^2=4(O{A}^2+O{B}^2)$.

Answer 1

Since, diagonals of a Rhombus intersect each other at right angels.

$\therefore$ in $\Delta AOB,$

$A{B}^2=O{A}^2+O{B}^2$ ........... $\left(1\right)$

and in $\Delta BOC,$

$B{C}^2=O{B}^2+O{C}^2$ ........... $\left(2\right)$

and in $\Delta COD,$

$C{D}^2=O{C}^2+O{D}^2$ ........... $\left(3\right)$

and in $\Delta AOD,$

$A{D}^2=O{A}^2+O{D}^2$ ........... $\left(4\right)$

Above results are obtained by using Pythagorous theorem.

Now, $\left(1\right)+\left(2\right)+\left(3\right)+\left(4\right)$ gives

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=2(O{A}^2+O{B}^2+O{C}^2+O{D}^2)$

$=4(O{A}^2+O{B}^2)$ ...... [Since, $OD=OB$ & $OA=OC$]

$\therefore A{B}^2+B{C}^2+C{D}^2+A{D}^2=4(O{A}^2+O{B}^2)$