Question

X is a point on the side BC of triangle ABC. 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 TX ^2 = TB* TC

Answer 1

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Answer 2

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

So,

$\sf(TX {}^{2} = TB \: × \: TC)$

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

USING PROPORTIONALITY THEOREM

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

$\begin{lgathered}\sf = (XN || AC) \\ \sf= (XN || CM)\end{lgathered}$

USING PROPORTIONALITY THEOREM

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

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

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

$\sf(TX {}^{2} = TB \: × \: TC)$

$\sf{Therefore ,PROVED}$