We have,PQRS as the given parallelogram, then PQ∥RS and PS∥QR.Since PQ∥RS and TS is a transversal, then∠PTS = ∠TSR (Alternate interior angles) .

....(1)Now, TS bisects ∠S, then∠PST = ∠TSR ........(2)From (1) and (2), we get∠PTS = ∠PSTIn

∆PTS,∠PTS = ∠PST (Proved above)

⇒PS = PT [Sides opposite to equal angles are equal] .........(3)

But PS = QR (opposite sides of parallelogram are equal) ........(4)

PT = TQ (As T is the mid point of PQ) ..........(5)

From (3),(4) and (5),

we getQR = TQIn ∆TQR,QR = TQ

(Proved above)⇒∠QTR = ∠QRT [angles opposite to equal sides are equal] .........

(6)Since PQ∥RS and TR is a transversal, then∠TRS = ∠QTR (Alternate interior angles) .........(7)

From (6) and (7),

we get∠QRT = ∠TRS⇒TR bisects ∠R.

Now, ∠R + ∠S = 180° [Adjacent angles of ∥gm are supplementary]

⇒12∠R + 12∠S =90°⇒∠TRS + ∠TSR = 90° .......(8)I

In ∆TSR,∠TRS + ∠TSR + ∠RTS = 180° [Angle sum property]⇒90° + ∠RTS = 180°⇒∠RTS = 90°

We have,PQRS as the given parallelogram, then PQ∥RS and PS∥QR.Since PQ∥RS and TS is a transversal, then∠PTS = ∠TSR (Alternate interior angles) .....(1)

Now, TS bisects ∠S, then

∠PST = ∠TSR ........(2)

From (1) and (2), we get∠PTS = ∠PSTIn

∆PTS,∠PTS = ∠PST (Proved above)

⇒PS = PT [Sides opposite to equal angles are equal] .........(3)

But PS = QR (opposite sides of parallelogram are equal) ........(4)

PT = TQ (As T is the mid point of PQ) ..........(5)

From (3),(4) and (5),

we getQR = TQIn ∆TQR,QR = TQ

(Proved above)

⇒∠QTR = ∠QRT [angles opposite to equal sides are equal] .........(6)

Since PQ∥RS and TR is a transversal, then

∠TRS = ∠QTR (Alternate interior angles) .........(7)

From (6) and (7),

we get∠QRT = ∠TRS⇒TR bisects ∠R.

Now, ∠R + ∠S = 180° [Adjacent angles of ∥gm are supplementary]

⇒12∠R + 12∠S =90°⇒∠TRS + ∠TSR = 90° .......(8)I

In ∆TSR,∠TRS + ∠TSR + ∠RTS = 180° [Angle sum property]

⇒90° + ∠RTS = 180°

⇒∠RTS = 90°