Question

$O$ is any point inside rectangle $ABCD$. Prove that $O{B}^2+O{D}^2=O{A}^2+O{C}^2$.

Answer 1

To prove that:

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

It can be easily proved that $APRD$ and $BPRC$ are rectangles

Hence, $AP=DR.........\left(i\right)$

$BP=CR.........\left(ii\right)$

Using pythagoras theorem,

$O{B}^2=B{P}^2+O{P}^2..........\left(iii\right)$

$O{D}^2=O{R}^2+R{D}^2........\left(iv\right)$

$O{A}^2=A{P}^2+O{P}^2.........\left(v\right)$

$O{C}^2=O{R}^2+R{C}^2.........\left(vi\right)$

LHS

$O{B}^2+O{D}^2=B{P}^2+O{P}^2+O{R}^2+D{R}^2$ using (iii) and (iv)

$=O{P}^2+O{R}^2+B{P}^2+A{P}^2$ using (i)

RHS

$O{A}^2+O{C}^2=A{P}^2+O{P}^2+O{R}^2+C{R}^2$ using (v) and (vi)

$=O{P}^2+O{R}^2+A{P}^2+B{P}^2$ using (ii)

LHS = RHS

Hence, proved