Theorem 5.12 Case 4

Case 4: B is outside of angle MON, and so is its reflection about OM, and O is inside angle BB"B'.

In this case, our previous considerations lead us to conclude that

angle BOB' = 2angle MON'

=2(180^o + angle MON )

= 360^o + 2angle MON

But adding 360^o is the same as adding nothing, so this is equal to

= 2angle MON

= angle AOA '

as in the other cases.