Solution:-

(i) From the figure, it is given that,

Consider the ∆ABC,

AD/DB = AE/EC

x/(x – 2) = (x + 2)/(x – 1)

By cross multiplication we get,

X(x – 1) = (x – 2) (x + 2)

x² – x = x² – 4

-x = -4

x = 4

(ii) From the question it is given that,

DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3

Consider the ∆ABC,

AD/DB = AE/EC

2x/(x – 2) = (2x + 3)/(x – 3)

By cross multiplication we get,

2x(x – 2) = (2x + 3) (x – 3)

2x² – 4x = 2x² – 6x + 3x – 9

2x² – 4x – 2x² + 6x – 3x = -9

-7x + 6x = -9

-x = – 9

x = 9