Question

Let, $X$ be any point on the side $BC$ of a triangle $ABC,$ if $XM$ & $XN$ are drawn parallel to $BA$ & $CA$ meeting $CA$ & $BA$ in $M$ & $N$, respectively. If $MN$ meets $CB$ produced in $T$. Then:

• A. $T{B}^2=TX\times TC$
• B. $T{C}^2=TB\times TX$
• C. $T{X}^2=TB\times TC$
• D. $T{X}^2=2(TB\times TC)$

Answer 1

The correct option is C $T{X}^2=TB\times TC$

Given: $XM||AB$ , $XN||AC$

$⟹BN||XM$ , $XN||CM$

Now, in $△TXM$ , we have, by Basic Proportionality theorem :

$\frac{TB}{TX}=\frac{TN}{TM}$ ------------(i)

Now, in $△TMC$ , we have, by Basic Proportionality theorem :

$\frac{TX}{TC}=\frac{TN}{TM}$ ------------(ii)

From (i) and (ii)

$\frac{TB}{TX}=\frac{TX}{TC}$

$⟹T{X}^2=TB\times TC$