A) 4 |
B) 2 |
C) 1 |
D) 8 |
A) 4 |
\( \Large \because DE \parallel BC \)
\( \Large \therefore \frac{AD}{DB} = \frac{AE}{EC} \)
=> \( \Large \frac{x}{x-2} = \frac{x+2}{x-1} \)
=> \( \Large x^{2} - x = x^{2} - 4 \)
=> x = 4