Question

If $ABCD$ is a rhombus then prove that ${AB}^2+{BC}^2+{CD}^2+{DA}^2={AC}^2+{BD}^2$.

Answer 1

Since $ABCD$ is a rhombus, therefore, its diagonal bisect each other at ${90}^{\circ }$

Hence, $\angle AOC=\angle BOC=\angle COD=\angle AOD={90}^{\circ }$.

Also, $AO=OC$ and $BO=OD$

Now, consider right $\Delta AOB$, By Pythagoras Theorem,

$\Rightarrow A{B}^2=A{O}^2+B{O}^2$

$\Rightarrow A{B}^2=(\frac{AC}{2}{)}^2+(\frac{BD}{2}{)}^2$

$\Rightarrow A{B}^2=\frac{A{C}^2}{4}+\frac{B{D}^2}{4}$

$\Rightarrow 4A{B}^2=A{C}^2+B{D}^2$...(i)

Similarly we can get,

$\Rightarrow 4B{C}^2=A{C}^2+B{D}^2$...(ii)

$\Rightarrow 4D{C}^2=A{C}^2+B{D}^2$...(ii)

$\Rightarrow 4A{D}^2=A{C}^2+B{D}^2$...(iv)

On adding all the equations, we get,

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

Hence, ${AB}^2+{BC}^2+{CD}^2+{DA}^2={AC}^2+{BD}^2$