Question

L is point on side QR of ΔPQR.If LM ||PR and LN||PQ and line MN meets the produced line QR at T, then prove that LTsq.=RT.TQ

Answer 1

GIVEN :PQR is a ∆, in which side QR is produced to point T.Also, LN∥PQ    and   LM∥PR.TO PROVE : LT2 = RT × TQPROOF :   In ∆ TQM, we have, LN∥QM, so,      LTQL = NTMN    By BPT theorem   ⇒QLLT = MNNT⇒QLLT + 1 = MNNT + 1     on adding '1' both sides⇒QL + LTLT = MN + NTNT⇒QTLT = MTNT     .........1In ∆ TLM, we have, LM∥RN, so,  RTLR = NTMN By BPT theorem  ⇒LRRT = MNNT⇒LRRT + 1 = MNNT + 1⇒LR + RTRT = MN + NTNT⇒LTRT = MTNT     ..........2So, from 1 and 2, we getQTLT = LTRT⇒LT2 = QT × RT

Answer 2

Answer:

GIVEN :PQR is a ∆, in which side QR is produced to point T.Also, LN∥PQ    and   LM∥PR.TO PROVE : LT2 = RT × TQPROOF :   In ∆ TQM, we have, LN∥QM, so,      LTQL = NTMN    By BPT theorem   ⇒QLLT = MNNT⇒QLLT + 1 = MNNT + 1     on adding '1' both sides⇒QL + LTLT = MN + NTNT⇒QTLT = MTNT     .........1In ∆ TLM, we have, LM∥RN, so,  RTLR = NTMN By BPT theorem  ⇒LRRT = MNNT⇒LRRT + 1 = MNNT + 1⇒LR + RTRT = MN + NTNT⇒LTRT = MTNT     ..........2So, from 1 and 2, we getQTLT = LTRT⇒LT2 = QT × RT

Step-by-step explanation: