Correct option is (D) All of these
Let diagonals of rhombus ABCD will intersect at O. Since, diagonals in a rhombus bisect each other at right angle.
\(\therefore OA=OC,OB=OD\;\&\;\angle AOB=90^\circ\)
\(\Rightarrow OA=\frac{AC}2\;\&\;OB=\frac{BD}2\)
Now in right \(\triangle AOB,\)
\(OA^2+OB^2=AB^2\) (By Pythagoras theorem)
\(\Rightarrow\) \((\frac{AC}2)^2+(\frac{BD}2)^2=AB^2\)
\(\Rightarrow\) \(AC^2+BD^2=4AB^2\)
\(=4BC^2\) \((\because BC=AB)\)
\(=4CA^2\) \((\because CA=AB)\)
