Question

PQRS IS A PARALLELOGRAM IN WHICH THE BISECTORS OF ANGLES OF P AND Q MEET ON RS. PROVE PQ = 2QR

Answer 1

Answer:

Since PQRS is a ∥gm, then PQ∥SR and  since  PQ∥SR and TQ is a transversal, then = ∠TQP   [Alternate interior angles]  (1)Since TQ bisects ∠Q,  ∠TQP    (2)From (1) and (2),  = Now, in ΔRTQ, we have  ∠RTQ = ∠TQR  (proved above)⇒QR = TR  [Sides opposite to equal angles in a Δ are equal]  (3)Now, since  PQ∥SR and TP is a transversal,  = ∠TPQ   [Alternate interior angles]  (4)Since TP bisects ∠P,  = ∠TPQ    (5)From (4) and (5), we get in ΔSTP, we have = ∠TPS    (proved above)⇒PS = ST  [Sides opposite to equal angles in a Δ are equal]  (6)Now, PS = QR (as, opposite sides of ∥gm are equal)⇒ST = TR   [using (3) and (6)]⇒T is the mid point of S SR = 2TR⇒PQ = 2 QR    [as from (3), TR = QR]

Step-by-step explanation:

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