Rules of Inscribed Angles

Question

https://www.dropbox.com/s/chbs2vilr9wjkvz/20140819_130744.jpg

Image of question found above.

I don't understand why angle BCD is formed by tangent and chord and is equal to 1/2 of arc BC.

Answer 1

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 $\angle BCD$

$\angle OCD = \angle OCB + \angle BCD = 90$