Question

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Answer 1

Hi mate here is the answer:-✍️✍️

❇️Question:-In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR =1/2 ∠QPR.

❇️ Solution:-In ΔQTR,

∠TRS = ∠TQR + ∠QTR (Exterior angle of a triangle equals to the sum of the two opposite interior angles)

∠QTR = ∠TRS - ∠TQR -----------(i)

Similarly in ΔQPR,

∠SRP = ∠QPR + ∠PQR

2∠TRS = ∠QPR + 2∠TQR (∵ ∠TRS and ∠TQR are the bisectors of ∠SRP and ∠PQR respectively.)

∠QPR = 2 ∠TRS - 2 ∠TQR

angleTRS- angleTQR.= 1/2 PQR............(ii)

From (i) and (ii), we get

angle QTR= 1/2 QPR

Hence, proved