In ∆ABC, D and E are point on AC and BC respectively such that DE║AB. If AD=2x, BE = 2x-1, CD= X+1 then find the value of x.
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you did not mentioned about CE,
substitute CE and simplify it
you can get. x
substitute CE and simplify it
you can get. x
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amankhanab111:
CE=x-1
(x+1)/2x =(x-1)/(2x-1)
(x+1)(2x-1)=2x(2x-1)
above line is wrong
(x+1)(2x-1)=2x(x-1)
2x^2-x+2x-1=2x^2-2x
after cancellation -x+2x-1=-2x remains
-x+2x+2x=1
3x=1
x=1/3
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