in a llgm PQRS . T is the midpoint of PQ and ST bisect angleS . provethat
Answers
Answer:
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°
Answer:
We have, PQRS as the given parallelogram, then PQIIRS and PSIIQR.Since PQIIRS 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
APTS, 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 ATQR,QR = TQ
(Proved above) <QTR = 2QRT [angles opposite to equal sides are equal].
(6)Since PQ||RS and TR is a transversal, then TRS QTR (Alternate interior angles)..
From (6) and (7),
we get QRT = <TRS TR bisects <R.
Now, ZR + <S = 180° [Adjacent angles of Ilgm are supplementary]
12/R+ 12/S =90° TRS + TSR =
90°.......(8)1
In ATSR,<TRS + <TSR + <RTS = 180° [Angle sum property] 90° + <RTS = =
180°⇒ <RTS = 90°