Question

In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS² + TQ² = TP² + TR² (As shown in the figure, draw seg AB || side SR and A-T-B)

Answer 1

According to Pythagoras theorem, in ∆PAT

${\mathrm{PT}}^2={\mathrm{PA}}^2+{\mathrm{AT}}^2...\left(1\right)$

In ∆ATS

${\mathrm{TS}}^2={\mathrm{AT}}^2+{\mathrm{AS}}^2...\left(2\right)$

In ∆QBT

${\mathrm{QT}}^2={\mathrm{QB}}^2+{\mathrm{BT}}^2...\left(3\right)$

In ∆BRT

${\mathrm{TR}}^2={\mathrm{BT}}^2+{\mathrm{BR}}^2...\left(4\right)$

Now,
$$\begin{gathered} {\mathrm{TS}}^2+{\mathrm{TQ}}^2=\left({\mathrm{AS}}^2+{\mathrm{AT}}^2\right)+\left({\mathrm{QB}}^2+{\mathrm{BT}}^2\right)\left(\mathrm{From}\left(2\right)\mathrm{and}\left(3\right)\right) \\ \Rightarrow {\mathrm{TS}}^2+{\mathrm{TQ}}^2=\left({\mathrm{BR}}^2+{\mathrm{AT}}^2\right)+\left({\mathrm{PA}}^2+{\mathrm{BT}}^2\right)\left(\because \mathrm{AS}=\mathrm{BR}\mathrm{and}\mathrm{PA}=\mathrm{QB}\right) \\ \Rightarrow {\mathrm{TS}}^2+{\mathrm{TQ}}^2=\left({\mathrm{BR}}^2+{\mathrm{BT}}^2\right)+\left({\mathrm{PA}}^2+{\mathrm{AT}}^2\right) \\ \Rightarrow {\mathrm{TS}}^2+{\mathrm{TQ}}^2=\left({\mathrm{TR}}^2\right)+\left({\mathrm{PT}}^2\right)\left(\mathrm{From}\left(1\right)\mathrm{and}\left(4\right)\right) \\ \Rightarrow {\mathrm{TS}}^2+{\mathrm{TQ}}^2={\mathrm{TR}}^2+{\mathrm{PT}}^2 \end{gathered}$$

Hence, TS² + TQ² = TP² + TR².