Given:

□PQRS is a rectangle

T is in the interior of □PQRS

To prove:

TS² + TQ² = TP² + TR²

Construction:

Draw a line parallel to side SR, through point T, intersecting sides PS and QR at point A and B such that (A – T – B).

Proof:

□PQRS is a rectangle …[Given]

∴ ∠SPQ = ∠PSR = ∠SRQ = ∠PQR = 90° …(i) [Each angle of a rectangle is 90⁰]

∴ seg PQ || seg SR …(ii) [Opposite sides of rectangle are parallel]

But, 

seg AB || seg SR …(iii) [Construction]

∴ seg PQ || seg SR || seg AB …(iv) [From (i), (ii) & (iii)] 

seg PQ || seg AB and PS is the transversal …[From (iv)]

∴ ∠QPS = ∠BAS = 90° …(v) [Corresponding angles test]

And,

seg PQ || seg AB and QR is the transversal …[From (iv)]

∴ ∠PQB = ∠ABR = 90° …(vi) [Corresponding angles test]

Now, 

∠BAS = 90° …[From (v)]

∴ ∠PAB = 90° …(vii) [Linear angles test]

Also,

∠ABR = 90° …[From (vi)]

∴ ∠ABQ = 90° …(viii) [Linear angles test]

In ∆PAT,

∠PAT = 90° …[From (vii), A – T – B]

∴  TP² = PA² + AT² …(ix) [By Pythagoras theorem]

In ∆SAT,

∠SAT = 90° …[From (v), A – T – B]

∴  TS² = AS² + AT² …(x) [By Pythagoras theorem]

In ∆QBT,

∠QBT = 90° …[From (viii), A – T – B]

∴  TQ² = QB² + BT² …(xi) [By Pythagoras theorem]

In ∆RBT,

∠RBT = 90° …[From (vi), A – T – B]

∴  TR² = BR² + BT² …(xii) [By Pythagoras theorem]

In □ABRS, 

seg AB || seg SR …[From (iii)]

seg AS || seg BR …[Opposite sides of rectangle are parallel and P – A – S, Q – B – R]

∴ □ABRS is parallelogram …[By definition]

∴ AS = BR …(xiii) [Opposite sides of parallelogram are equal]

And,

In □ ABQP,

seg AB || seg PQ …[From (iv)]

seg AP || seg BQ …[Opposite sides of rectangle are parallel and P – A – S, Q – B – R]

∴ □ABPQ is parallelogram …[By definition]

∴ AP = QB …(xiv) [Opposite sides of parallelogram are equal]

Adding (x) and (xi), we get,

TS² + TQ² = AS² + AT² + QB² + BT²

∴ TS² + TQ² = BR² + AT² + AP² + BT²

∴ TS² + TQ² = BR² + BT² + AP² + AT²

∴ TS² + TQ² = TR² + TP² …[From (ix) and (xii)]

Hence proved.