Question

Diagonals of rhombus $ABCD$ intersect each other at point $O$. Prove that:
$O{A}^2+O{C}^2=2A{D}^2-\frac{B{D}^2}{2}$

Answer 1

Diagonals of a rhombus intersect at right angle.

Thus, the triangles formed are right angled triangle.

In triangle $AOD,\angle AOD={90}^o$

$\therefore O{A}^2=A{D}^2-O{D}^2.................\left(1\right)$

In triangle $OBC,$

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

Add equation $\left(1\right)$ and $\left(2\right),$ we get

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

$=2(AD{)}^2-O{D}^2-O{B}^2$ ..... $(\because AD=BC)$

$=2(AD{)}^2-{\left(\frac{BD}{2}\right)}^2-{\left(\frac{BD}{2}\right)}^2$

(Because $O$ is the mid point of $BD$)

$=2(AD{)}^2-\left(\frac{B{D}^2}{4}\right)-\left(\frac{B{D}^2}{4}\right)$

$=2A{D}^2-\left(\frac{B{D}^2}{2}\right)$

Hence proved.