Question

In ABC, L and M are the points on the
sides CA and CB such that LM is parallel
to AB. IF AL=X-3, AC= 2x, BM = x-2
BC2x +3 , find the value of x.

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Answer 1

Correct Question:-

In ∆ABC, L and M are the points on the sides CA and CB such that LM is parallel to AB. If AL = x - 3, AC = 2x, BM = x - 2 and BC = 2x + 3, find the value of x.

Solution:-

Given:-

• LM || AB
• AL = x - 3
• AC = 2x
• BM = x - 2
• BC = 2x + 3

To find:-

The value of x.

Note:-

Refer to the attachment for a clear concept.

In triangle:-

As the L and M are points on AC and BC respectively,

LC + AL = AC

$\implies$ LC = AC - AL $\longrightarrow$ (1)

MC + BM = BC

$\implies$ MC = BC - BM $\longrightarrow$ (2)

Solution:-

In ∆ABC,

LM || AB

According to Thale's Theorem,

$\sf{\dfrac{AL}{LC} = \dfrac{BM}{MC}}$

Now Substituting the values of LC and MC from (1) and (2)

= $\sf{\dfrac{AL}{AC - AL} = \dfrac{BM}{BC - BM}}$

Now, Substituting the values from given,

$\sf{\dfrac{x-3}{2x - (x-3)} = \dfrac{x-2}{(2x+3)-(x-2)}}$

= $\sf{\dfrac{x-3}{2x-x+3} = \dfrac{x-2}{2x+3 -x+2}}$

= $\sf{\dfrac{x-3}{x+3} = \dfrac{x-2}{x+5}}$

By Cross-multiplication,

$\sf{(x-3)(x+5) = (x+3)(x-2)}$

= $\sf{x(x+5) -3(x+5) = x(x-2)+3(x-2)}$

= $\sf{x^2 + 5x - 3x -15 = x^2 - 2x +3x - 6}$

= $\sf{x^2 + 2x -15 = x^2 + x - 6}$

Taking all the variables on LHS,

The signs of the variables will change according to:-

(-) into (+)

(+) into (-)

= $\sf{x^2 + 2x -15 - x^2 - x + 6 = 0}$

Taking like terms together.

= $\sf{x^2 - x^2 + 2x - x - 15 + 6 = 0}$

= $\sf{\cancel{x^2} - \cancel{x^2} + x -9 = 0}$

= $\sf{x - 9 = 0}$

= $\sf{ x = 9}$

Therefore the value of x is 9

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Additional Information:-

→ What is Thale's Theorem?

✓ Thale's Theorem states that if a line which is parallel to one side of the triangle intersecting the other two at distinct points then the line divides the two sides in proportion.

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