The value of x-2
de is
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Answer:
Given: ABC is a triangle, DE || BC, AD = x, DB = x - 2, AE = x + 2 and EC = x - 1.
To find: x
In △ABC, we have
DE || BC
Therefore [By Thale's theorem]
AD/DB = AE/EC
AD × EC × = AE × DB
x(x-1) = (x-2)(x+2)
x2 - x = x2 - 4
x = 4
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