Q19 of 56 Page 6

In fig. DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value x.                         
         

Given: ABC is a triangle, DE || BC, AD = x, DB = x – 2, AE = x + 2 and EC = x – 1.
To find: x


In Δ ABC, we have
DE || BC
Therefore [By Thale's theorem]

AD × EC = AE × DB
x(x - 1) = (x - 2)(x + 2)
x2 – x = x2 – 4
x = 4		   



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