Question

The diagonals of a convex $\square PQRS$ intersect at right angles then prove that
$P{Q}^2+R{S}^2=P{S}^2+Q{R}^2$

Answer 1

Given: The diagonals of a convex $\square ABCD$ intersect at right angles at $O$.
To Prove: $P{Q}^2+R{S}^2=P{S}^2+Q{R}^2$
Proof: In $\Delta POS,m\angle O={90}^o$
$\therefore P{S}^2=O{P}^2+O{S}^2$ ......... (i)
In $\Delta QOR,m\angle O={90}^o$
$\therefore Q{R}^2=O{Q}^2+O{R}^2$ ......... (ii)
In $\Delta POQ,m\angle O={90}^o$
$\therefore P{Q}^2=O{P}^2+O{Q}^2$ ......... (iii)
In $\Delta ROQ,m\angle O={90}^o$
$\therefore Q{R}^2=O{R}^2+O{S}^2$ ........ (iv)
Adding results (3) and (4)
$P{Q}^2+Q{R}^2=O{P}^2+O{Q}^2+O{R}^2+O{S}^2$
$\therefore P{Q}^2+Q{R}^2=(O{P}^2+O{S}^2)+(O{Q}^2+O{R}^2)$
$\therefore P{Q}^2+R{S}^2=P{S}^2+Q{R}^2$ ......... [from $\left(i\right)$ and $\left(ii\right)$]