In a triangle ABC, D and E are points on sides AB and AC respectively such that DE || BC. If AD = x, DB = x-2, AE = x+2, and EC = x-1, find the value of x.

Solution

Solution: Since DE is parallel to BC, by the Basic Proportionality Theorem, \(\frac{AD}{DB}=\frac{AE}{EC}\). Substituting, \(\frac{x}{x-2}=\frac{x+2}{x-1}\). Cross-multiplying gives \(x(x-1)=(x-2)(x+2)=x^2-4\). Thus, \(x^2-x=x^2-4\Rightarrow x=4\). Trick: For a line parallel to one side of a triangle, the corresponding segments on the other two sides are proportional.