Question

Ab is a diameter of the circle cd is a chord equal to the radius of the circle. Ac and bd when extended intrsect at a point e

Answer 1

Hope this helps you buddy ☺️☺️⭐✨✨⭐✌️✌️

Answer 2

AnswEr:

$\bf\underline{Solution:-}$ Join OC, OD and BC.

In triangle OCD, we have

OC = OD = CD $\qquad\sf{[Each\:equal\:to\:radius]}$

$\Rightarrow$ $\sf{\triangle\:OCD\:is\: equilateral}$

$\Rightarrow$ $\sf\underline{\angle\:COD\:=60°}$

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Now, $\sf{\angle\:CBD=}$$\sf\dfrac{1}{2}\angle\:COD$

$\Rightarrow$ $\angle$$\sf{CBD=30°}$

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Since $\angle$ACB is angle in a semi-circle.

$\therefore$ $\angle$ ACB = 90°

$\rightarrow$ $\angle$BCE = 180° - $\angle$ACB = 180° - 90° = 90°.

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$\star$ $\sf\underline{Thus,\:in\triangle\:BCE,\:we\:have}$

$\angle$BCE = 90° and $\angle$CBE = $\angle$CBD = 30°

$\therefore$ $\angle$BCE + $\angle$CBE + $\angle$CEB = 180°.

$\Rightarrow$ 90° + 30° + $\angle$CEB = 180°.

$\Rightarrow$ $\angle$ CEB = 60°.

$\Rightarrow$ $\angle$ AEB = 60°.

#BAL

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