Question

Let X be any point on side BC of $\Delta ABC$. seg $XM||sideAB$ and $segXN||sideCA$. Prove that: $T{X}^2=TB.TC.$

Answer 1

Let $X$ be any point on side $AB$.

Given that, $XM\parallel ABandXN\parallel AC$

To prove that, $T{X}^2=TB.TC$

Proof:- $BN\parallel XM$

Now,

In $\Delta TXM$,By using Basic proportionality theorem, we have

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

Now,

In $\Delta TCM$, By using Basic proportionality theorem, we have

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

By equation (1) and (2) to, we get

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

$\Rightarrow TB.TC=T{X}^2$

$\Rightarrow T{X}^2=TB.TC$

Hence proved.