Question

In triangle ABC DE parallel to BC, if AD=x, DB=x-2,AE=X+2,EC=X-1 find the value of x
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Answer 1

Answer:

By B.P.T.,

AD/DB=AE/EC

x/x-2=x+2/x-1

so

x(x-1)=(x-2)(x+2)

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

cancelling

${x}^{2} on both sides$

so

x=4

Answer 2

Given, DE//BC

$\frac{AD}{DB} = \frac{AE}{EC} \\ \\ \frac{x}{x -2 } = \frac{x + 2}{x - 1} \\ \\ x(x - 1) = (x + 2)(x - 2) \\ \\ {x}^{2} + x = {x}^{2} + 4 \\ \\ x = 4$