In the adjoining figure, $$BOC$$ is the diameter of the semicircle with centre O. DE is the tangent at D . If $$AB = k(AE)$$, then the numerical value of $$k$$ is

Since $$AB$$ is perpendicular to the diameter at $$B$$ hence it is also a tangent at $$B$$

Now the two tangents from the same point to the circle are equal which implies

$$EB = ED$$ and $$\angle EDO = 90\degree$$

Consider $$\triangle EDO, \triangle EBO$$

$$ED = EB, BO =DO = r, EO = EO$$

Thus, the triangles are congruent to each other

$$\angle EOB = \angle EOD = x$$

Consider $$\triangle DOC$$

$$\angle DOC = 180 -2x$$

$$ CO = DO = r$$. 

Thus, it's an isosceles triangle

$$\angle DCO=\dfrac{1}{2}(180-(180-2x))=x$$

Now, consider $$\triangle EOB, \triangle ACB$$

$$\angle ACB = \angle EOB = x$$

$$\angle ABC = \angle EBO = 90\degree$$

Thus, by AA similarity criterion, the triangles are similar to each other

$$ \dfrac{AE}{AB} = \dfrac{BO}{BC} = \dfrac{r}{2r} = \dfrac{1}{2}$$

$$\implies AB = 2AE$$

$$ k = 2$$