Find ∠BDE in a Circle: BDC is a Minor Arc with ∠BOC = 110°

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To find the angle ∠BDE in a circle where BDC is a minor arc and ∠BOC = 110°, it is established that ∠BDE measures 55 degrees, which is exactly half of ∠BOC. This conclusion is drawn from the relationship between angles subtended by arcs in a circle. The discussion also references the theorem stating that ∠BOC = 2*∠BPC, indicating the need for additional theorems related to cyclic quadrilaterals to fully understand the relationships between these angles.

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1. BDC is a minor arc of a circle, centre O. CD is produced to E. If ∠BOC = 110°, find the number of degrees in ∠BDE.

this is the diagram i came up with,

http://img209.imageshack.us/img209/2500/circle4hm.jpg

by experimentation (i know it's weak), i found ∠BDE to be 55, exactly half of ∠BOC and i can't find a theorem to relate to this. anyone have ideas?

thanks.
 
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If you draw a point P on the major arc of OBC, then there is a theroem which states that: ∠BOC = 2*∠BPC.
You still need another theroem though, about the opposite angles of a cyclic quadrilateral. Can you find that?
 

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