Question

Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA, meeting CA, BA in M, N respectively. M and N are joined and produced to meet the line passing through BC at T as shown in the figure. Prove that ${TX}^2=TB\times TC$.

Answer 1

In $△TXM$, we have

$XM||BN$ {$\because XM||AB$}

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

In $△TMC$, we have

$XN||CM$ {$\because XN||AC$}

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

From equation (i) and (ii), we get

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

$\Rightarrow$ ${TX}^2=TB\times TC$ (hence proved)