Question

In parallelogram PQRS, T is midpoint of PQ and ST bisects angle S. Prove that
a) QR=QT
b) RT bisects Angle R
c) Angle STR = 90'. Give reasons.​

Answer 1

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Answer 2

a) QR=QT

Sol :-

∠PST = ∠TSR - - - - (1)

∠PTS = ∠TSR - - - - - (2) {alternate angle ∵ Sr || PQ}

From (i) and (ii)

∠PST = ∠PTS

Therefore,

PT = PS

But PT = QT ...(T is midpoint of PQ)

And PS = QR ... (PS and QR are opposite and equal sides of a parallelogram)

Hence,

QT = QR.

b) RT bisects Angle R

Sol :-

Since QT = QR

∠QTR = ∠QRT

But ∠QTR = ∠TRS. - - - (Alternate angles ∵ SR || PQ)

=> ∠QRT = ∠TRS

Therefore, RT bisects ∠R

c) Angle STR = 90°

Sol :-

∠PST = ∠TSR

∠QRS = ∠TRS

∠QRS + ∠PSR = 180° - - (adjacent angle of ||gm are supplementary)

$\sf \: Multiplying \: by \: \frac{1}{2}$

$\frac{1}{2} \angle \: QRS \: + \frac{1}{2} \angle \:PSR = \frac{1}{2} \times x18 {0}^{0}$

In Δ RTS,

∠TSR + ∠RTS + ∠TRS = 180°

90° + ∠RTS = 180°

∠RTS = 180 - 90

∠RTS = 90°