A) 3
B) 4
C) 5
D) 6
Correct Answer: B
Solution :
[b] \[\frac{AD}{DB}=\frac{AE}{EC}\] as \[DE\,\,||BC\]
\[\therefore \,\,\,\frac{x}{x-2}=\frac{x+2}{x-1}\] or \[{{x}^{2}}-\text{ }x={{x}^{2}}-4\] Which gives x = 4
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