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I don't understand why angle BCD is formed by tangent and chord and is equal to 1/2 of arc BC.
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Notice that $\angle OCD = 90 $ as radius $\perp$ tangent
In isosceles $\triangle OCB$ : $\angle OCB = \dfrac{180 - \angle COB}{2} = 90 - \dfrac{\angle COB}{2} $
Next, use below to solve $\angle BCD$
$\angle OCD = \angle OCB + \angle BCD = 90$
AgentS
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The problem says that angle BCD=40. – Aug 19 '14 at 21:00
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Ahh okay so you're trying to find $\angle A$,are you familiar with inscrbed angle formula : $\text{inscribed angle} = \dfrac{\text{central angle }}{2}$ ? – AgentS Aug 19 '14 at 21:04
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no, I am not familiar. What would the central angle be? – Aug 19 '14 at 21:06
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for this particular problem, $\angle A = \dfrac{\angle COB}{2} = \angle BCD = 40^{\circ}$ – AgentS Aug 19 '14 at 21:08
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"central angle" is angle intercepted by arc at the center of circle. see http://www.geom.uiuc.edu/~dwiggins/pict44.GIF – AgentS Aug 19 '14 at 21:09
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doesn't angle COB=90 degrees? – Aug 19 '14 at 21:10
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nope, $\angle OCD$ equals 90 degrees – AgentS Aug 19 '14 at 21:11
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how do you solve for COB? – Aug 19 '14 at 21:12
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$\angle COB$ is the "central angle" for the "inscribed angle" $\angle A$. You knew $\angle A = 40$ , so use the inscribed angle formula – AgentS Aug 19 '14 at 21:14
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but I don't know what A equals. finding A is the question. – Aug 19 '14 at 21:17
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Ohhk.. first find angle OCB using : $ \angle OCB + \angle BCD = 90 $ – AgentS Aug 19 '14 at 21:19
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OCB=50 degrees then what? – Aug 19 '14 at 21:20
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Find angle COB using the triangle sum property : $\angle OCB + \angle OBC + \angle COB = 180^{\circ}$ – AgentS Aug 19 '14 at 21:22
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Note that the triangle is isosceles, so $\angle OCB = \angle OBC = 50$ this gives, $50 + 50 + \angle COB = 180$. solve $\angle COB$ – AgentS Aug 19 '14 at 21:23
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Why didn't I think of triangle sum property? – Aug 19 '14 at 21:24
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Haha happens :) solving we get : $\angle COB = 80$. appealing to inscribed angle formula gives $\angle A = \dfrac{\angle COB}{2} = \dfrac{80}{2} = 40$ – AgentS Aug 19 '14 at 21:26
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It is a good formula to memorize : $\angle BCD = \dfrac{\angle COB}{2}$. This works for ALL chord-tangent intersections – AgentS Aug 19 '14 at 21:27