Question

in the figure, □BCD is a parallelogram . T is any point on the side PQ. show that A (triangle TQS)= A (triangle TOR)​

Answer 1

Answer:

Join A and C

2. Let the point of intersection of AC and BD be 'O'.

Centroid is the point of intersection of medians of a triangle. AC and BD bisect each other (parallelogram property) and hence AO and CO are medians of triangle ABD and BCD respectively.

AO = CO - parallelogram property

Property: Centroid divides the median in the ratio 2:1.

Thus,

AE = 2EO

CX = 2XO

AO = CO

ie.

AE = CX and EO = OX

EX = EO + OX = 2EO = AE

Hence Proved.

Answer 2

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°