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 $$B$$, it is a tangent at $$B$$, so the two tangents from $$E$$ give $$EB = ED$$ and hence $$\angle EBD = \angle EDB$$. Because $$BOC$$ is a diameter, $$\angle BDC = 90^\circ$$, so $$\angle ADB = 90^\circ$$ and $$\angle ADE = 90^\circ - \angle EDB = 90^\circ - \angle ABD = \angle DAE$$. Therefore $$EA = ED = EB$$, making $$E$$ the midpoint of $$AB$$, so $$AB = 2(AE)$$ and $$k = 2$$.