LM || AB if AL = …
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Sia ? 6 years, 3 months ago
We have, AL = x - 3, AC = 2x, BM = x - 2 and BC = 2x + 3, and we need to find the value of x.
In {tex}\Delta{/tex}ABC, we have
{tex}L M \| A B{/tex}
{tex}\therefore \quad \frac { A L } { L C } = \frac { B M } { M C }{/tex} [By Thaley's Theorem]
{tex}\Rightarrow \quad \frac { A L } { A C - A L } = \frac { B M } { B C - B M }{/tex}
{tex}\Rightarrow \quad \frac { x - 3 } { 2 x - ( x - 3 ) } = \frac { x - 2 } { ( 2 x + 3 ) - ( x - 2 ) }{/tex}
{tex}\Rightarrow \quad \frac { x - 3 } { x + 3 } = \frac { x - 2 } { x + 5 }{/tex}
{tex} \Rightarrow{/tex} (x - 3) (x + 5) = (x - 2) (x + 3)
{tex} \Rightarrow{/tex} x2 + 2x -15 = x2 + x - 6
{tex} \Rightarrow{/tex} x = 9
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