Question

X is a point on the side bc of abc. Xm and xn are drawn parallel to ab and acrespectively meeting ab in n and ac in m. Mn produced meets cb produced at t.Prove that tx2= tb tc

Answer 1

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

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

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)]

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

$\sf(TX {}^{2} = TB \: × \: TC)$

$\sf{Therefore ,PROVED}$

Answer 2

$\bf{\huge{\boxed{\underline{\mathfrak{\red{Answer:}}}}}}$

Given : XM || AB, XN || AC

To Prove : TX2 = TB × TC

Proof: ∵ BN || XM

Now, in ∆TXM, we have

BN || XM.

Therefore, by using Basic proportionality theorem, we have

$\frac{TB}{TX} = \frac{TN}{TM} \: \: \: \: ............(1)$

And, XN || AC

XN || CM

Now, In ∆TMC, we have

∵ XN || CM

Therefore, By using Basic proportionality theorem, we have

$\frac{TX}{TC} = \frac{TN}{TM} \: \: ...............(2)$

Comparing (i) and (ii), we have

$\frac{tb}{tc} = \frac{TX}{TC} \\ \\ \huge \bold{ {TX}^{2} = TB \: \times \: TC}$