Question

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer 1

$\square ABCD$ is a rhombus with $O$ as point of intersection of diagonals.

In $\Delta AOB,\angle AOB={90}^0$ (since diagonals are perpendicular in rhombus).

By Pythagoras theorem,

${AB}^2={AO}^2+{OB}^2$

Similarly,

${BC}^2={OC}^2+{OB}^2,{DC}^2={OD}^2+{OC}^2{DA}^2={DO}^2+{OA}^2{AB}^2+{BC}^2+{CD}^2+{DA}^2=2({OA}^2+O{B}^2+{OC}^2+O{D}^2=4({AO}^2+D{O}^2)$

Rhombus diagonal biset each other,

$AO=OC,DO=OB$

$AC=AO+OC$

${AC}^2={OA}^2+O{C}^2+2AO.OC=4{AO}^2$

Similarly,

${DB}^2={4OD}^2\therefore {AC}^2+{DB}^2=4({AO}^2+D{O}^2){AB}^2+{BC}^2+{CD}^2+{DA}^2={AC}^2+{DB}^2$

Hence Proved.