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Circle Geometry Help (1 Viewer)

K-dogg

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From the diagram setup it seems likely you'll be using the alternate segment theorem.
I'll start from BDA = MAB (AST)
 

pikachu975

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let MAB = a, BAD = b
BDA = MAB = a (angles in alternate segments)
BCA = BDA = a (angles in the same segment)
NAC = BCA = a (alternate angles, MN || BC)
CDA = NAC = a (angles in alternate segments)

BDA = CDA = a therefore AD bisects BDC
 

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