Question

In the given figure $LM||AB$. If AL $=x-3$, AC $=2x$, BM$=x-2$ and BC $=2x+3$, find the value of $x$.

Answer 1

In $△ABC$, we have

$LM||AB$

$\therefore$ $\frac{AL}{LC}=\frac{BM}{MC}$ [By Thaley's Theorem]

$\Rightarrow$ $\frac{AL}{AC-AL}=\frac{BM}{BC-BM}$

$\Rightarrow$ $\frac{x-3}{2x-(x-3)}=\frac{x-2}{(2x+3)-(x-2)}$

$\Rightarrow$ $\frac{x-3}{x+3}=\frac{x-2}{x+5}$

$\Rightarrow$ $(x-3)(x+5)=(x-2)(x+3)$ [on cross multiplication]

$\Rightarrow$ $x^2+2x-15=x^2+x-6$

$\Rightarrow$ $x=9$ [cancelling $x^2$ from both sides and solving for $x$]