Question

In parallelogram PQRS, point E is the mid point of side QR. If SE and PQ are extend
on F, then prove that PF = 2PQ.
If you give wrong answer I will report it.
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Answer 1

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°

hope it helps you ....