At end A of a chord AB of a circle tangent is drawn. A point D is taken on tangent such that $\displaystyle \angle ABE=\angle ABD$ where E is some point on the circumference If BE = $3$ and BD = $6$ Find AD

Since angle ABE equals angle ABD, BE acts as the diameter of the circle, making angle BAE 90 degrees, which means angle DAE is also 90 degrees. By the tangent property, angle DAB is 90 degrees, so triangles DAE and DBA are similar. From the similarity, AD/BD = AE/AB and by the tangent secant theorem AD^2 = AE * AB; since BE is the diameter and triangle BAE is a right triangle, using AE = sqrt(BE^2 - AB^2) gives AD^2 = AE^2, leading to AD = 3 * sqrt(2).

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