(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.