In the figure given alongside DE \\ BC, AD=x , BD=x-2,
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