Question

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

Answer 1

$Inrhombus,diagonalbisectseachotherat{90}^{\circ }$

$\therefore \angle AOB=\angle BOC=\angle COD=\angle DOA={90}^{\circ }$

$Also,AO=CO=\frac{1}{2}AC\to \left(i\right)andBO=DO=\frac{1}{2}BD\to \left(ii\right)$

$In\Delta AOB,A{B}^2=O{A}^2+O{B}^2$

$={\left(\frac{1}{2}AC\right)}^2+{\left(\frac{1}{2}BD\right)}^2$

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

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

$A{B}^2+B{C}^2+C{D}^2+D{C}^2=A{C}^2+B{D}^{2}proved.$