Question

In AABC, DE || BC. AD = x, DB = x-2, AE = x + 2 and EC = x - 1. Find the value of x.​

Answer 1

Given,

$In \: \triangle \: ABC \: DE \: \parallel \: BC \: .AD = x, DB = x-2, AE = x + 2 \: and \: EC = x - 1. \:$

To find out,

The value of X.

Solution:

Basic proportionality theorem (Thales theorem):

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Here, DE\\BC.

So,

$\frac{AD}{BD} = \frac{AE}{CE} (by \: basic \: proportionality \: theorem.)$

$\frac{x}{x - 2} = \frac{x + 2}{x - 1}$

By cross multiplication.

$(x - 2)(x + 2) = (x)(x - 1) \:$

${x}^{2} - {2}^{2} = { {x}^{2} - x }$

${x}^{2} - 4 = { {x}^{2} - x }$

$- 4 = - x$

$4 = x$

$x = 4$

Therefore the value of x is 4.

Verification:

$(x - 2)(x + 2) = (x)(x - 1) \:$

$(4 - 2)(4 + 2) = (4)(4 - 1)$

$(2)(6) = (4)(3)$

$12 = 12$

L.H.S = R.H.S