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$\mathrm{AB}$ is a diameter of a circle. $\mathrm{CD}$ is a chord parallel to $\mathrm{AB }$ and $\mathrm{2CD = AB}$ . The tangent at $\mathrm{B}$ meets $\mathrm{AC}$ produced at $\mathrm{E.}$ Then $\mathrm{AE.}$ is equal to

Option: 1
$\mathrm{AB}$

Option: 2
$\mathrm{\frac{1}{2}AB}$

Option: 3
$\mathrm{4AB}$

Option: 4
$\mathrm{2AB}$

Answers (1)

best_answer

Clearly, $\mathrm{O C D}$ is equilateral triangle

So, $\mathrm{\angle C O D=60^{\circ}}$

Now, $\mathrm{O A=O C}$ and $\mathrm{\angle A O C=\angle B O D=60^{\circ}}$

So, $\mathrm{\angle A=60^{\circ}}$
Now, $\mathrm{ \cos 60^{\circ}=\frac{A B}{A E} \Rightarrow A E=2 A B }$

Hence option 4 is correct.

Posted by

Shailly goel

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