$AB$ is a diameter of a circle. $CD$ is a chord parallel to $AB$ and $2CD = AB$. The tangent at $B$

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$AB$ is a diameter of a circle. $CD$ is a chord parallel to $AB$ and $2CD = AB$. The tangent at $B$ meets the line $AC$ produced at $E$ then $AE$ is equal to:

Accepted Answer

[Key Concept: $AB$ a diameter, $CD \parallel \ AB$ with $2CD = AB$: use the inscribed-angle / tangent geometry.]From figure it is clear that$$\Rightarrow AE = \frac{AB}{cos60^{\circ}} = 2(AB) $$

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