Question

In the figure point $T$ is in the interior of rectangle $PQRS$
Prove that $T{S}^2+{TQ}^2={TP}^2+{TR}^2$
(As shown in the figure, draw seg $AB\parallel$ side $SR$ and $A-T-B$)

Answer 1

$PS=QR$ [ Opposite sides of rectangle are equal ]

$\Rightarrow$ $AS\parallel BR$ [ $P-A-S$ and $Q-B-R$ ]

$\Rightarrow$ Also, $AB\parallel SR$ [ Construction ]

$\therefore$ $\square ASRB$ is a parallelogram .

$\Rightarrow$ $\angle S={90}^o$ [ Angles of rectangle $PSRQ$ ]

A parallelogram is a rectangle , if one of its angles is a right angle.

$\therefore$ $ASRB$ is a rectangle

$\Rightarrow$ $\angle SAB=\angle ABR={90}^o$ [ Angles of a rectangle ]

$\therefore$ $TA\perp PS$ and $TB\perp QR$ ---- ( 1 )

$\Rightarrow$ $AS=BR$ ----- ( 2 ) [ Opposite sides of rectangle are equal ]

Similarly, we can prove $AP=BQ$ ----- ( 3 )

In n$△TAS$

$\Rightarrow$ $\angle TAS={90}^o$ [ From ( 1 ) ]

$\Rightarrow$ $T{S}^2=T{A}^2+A{S}^2$ ----- ( 4 ) [ By Pythagoras theorem ]

In $△TBQ,$

$\Rightarrow$ $\angle TBQ={90}^o$ [ From ( 1 ) ]

$\therefore$ $T{Q}^2=T{B}^2+B{Q}^2$ ----- ( 5 )

Adding ( 4 ) and ( 5 ) we get,

$T{S}^2+T{Q}^2=T{A}^2+A{S}^2+T{B}^2+B{Q}^2$ ------ ( 6 )

In $△TAP,$

$\Rightarrow$ $\angle TAP={90}^o$ [ From ( 1 ) ]

$\therefore$ $T{P}^2=T{A}^2+A{P}^2$ ------ ( 7 )

In $△TBR,$

$\Rightarrow$ $\angle TBR={90}^o$

$\therefore$ $T{R}^2=T{B}^2+B{R}^2$ ----- ( 8 )

Adding ( 7 ) and ( 8 ), we get,

$\Rightarrow$ $T{P}^2+T{R}^2=T{A}^2+A{P}^2+T{B}^2+B{R}^2$

$\therefore$ $T{P}^2+T{R}^2=T{A}^2+B{Q}^2+T{B}^2+A{S}^2$ [ From ( 2 ) and ( 3 )] ---- ( 9 )

From ( 6 ) and ( 9 ) we get,

$\Rightarrow$ $T{P}^2+T{R}^2=T{S}^2+T{Q}^2$