Question

X is a point on the side bc of 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 tx2 = tb tc

Answer 1

using similarity criteria of triangles we get the required relation

Answer 2

$\huge{\mathfrak{\green{S}}}{\mathfrak{o}}{\mathfrak {\orange {l}}}{\mathfrak{\red{u}}}{\mathfrak {\pink {t}}}{\mathfrak{\blue{i}}}{\mathfrak{\purple {o}}}{\mathfrak {\gray {n}}}$

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

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)]

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

$\sf(TX {}^{2} = TB \: × \: TC)$

$\sf{Therefore ,PROVED}$