Chapter 6 Class 9 Lines and Angles (Term 1)

Serial order wise

Last updated at March 16, 2017 by Teachoo

Ex 6.3 ,6 In the given figure, 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 Given TQ is the bisector of ∠ PQR. So, ∠ PQT = ∠ TQR = 1/2 ∠ PQR Also, TR is the bisector of ∠ PRS So, ∠ PRT = ∠ TRS = 1/2 ∠ PRS In Δ PQR, ∠ PRS is the external angle ∠ PRS = ∠ QPR + ∠ PQR In Δ TQR, ∠ TRS is the external angle ∠ TRS = ∠ TQR + ∠ QTR Putting ∠ TRS = 1/2 ∠ PRS & ∠ TQR = 1/2 ∠ PQR 1/2 ∠PRS = 1/2 ∠ PQR + ∠ QTR 1/2 ∠PRS = 1/2 ∠ PQR + ∠ QTR Putting ∠ PRS = ∠ QPR + ∠ PQR from (1) 1/2 (∠ QPR + ∠ PQR) = 1/2 ∠ PQR + ∠ QTR 1/2 ∠ QPR + 1/2∠ PQR = 1/2 ∠ PQR + ∠ QTR 1/2 ∠ QPR + 1/2∠ PQR – 1/2 ∠ PQR = ∠ QTR 1/2 ∠ QPR = ∠ QTR ∠ QTR = 1/2 ∠ QPR Hence proved