Question

X is a point on the side BC of ΔABC. XM and XN are drawn parallel to AB and AC respectively meeting AB in N and AC in M. MN produced meets CB produced at T. Prove that TX2 = TB × TC

Answer 1

∆ΤXN ~ ∆TCM

⇒ TX / TC = XN / CM = TN / TM

⇒ TX × TM = TC × TN ....(i)

Again, ∆TBN ~ ∆TXM

⇒ TB / TX = BN / XM = TN / TM

⇒ TM = (TN × TX) / TB ...(ii)

Using (ii) in (i), we get

TX2 × TN/TB = TC × TN

⇒ TX2 = TC × TB