Question

Let x be any point on side bc of a triangle abc. If xm and xn are drawn parallel to ba and ca meeting ca and ba at m and n respectively and mn meets bc produced at t prove that tx^2=tbxtx

Answer 1

Answer:

Step-by-step explanation:

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 .........(1)In ∆ TLM, we have, LM∥RN, so, RTLR = NTMN (By BPT theorem) ⇒LRRT = MNNT⇒LRRT + 1 = MNNT + 1⇒LR + RTRT = MN + NTNT⇒LTRT = MTNT ..........(2)So, from (1) and (2), we getQTLT = LTRT⇒LT2 = QT × RT

Answer 2

Answer:

Step-by-step explanation:

Hope it helps