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

Answer:

∆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

Answer 2

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$\sf(XM = AB, XN = AC)$ [Parallel]

So,

$\sf(TX {}^{2} = TB \: × \: TC)$

$\sf(BN || XM) = > parallel$

TXM

$\sf= (BN || XM) = > parallel$

USING PROPORTIONALITY THEOREM

$\sf( \frac{tb}{tx} = \frac{tn}{tm} )....eq - (1)$

$\sf = (XN || AC) \\ </h3><p> \sf= (XN || CM)$

TMC

$\sf = (XN || CM)$

USING PROPORTIONALITY THEOREM

$\sf( \frac{tx}{tc} = \frac{tn}{tm} ).....eq - (2)$

Comparing both equation[ eq - (1) and eq - (2)]

$\sf (\frac{tb}{tx} = \frac{tx}{tb})$

$\sf(TX {}^{2} = TB \: × \: TC)$

$\sf{Therefore ,PROVED}$