"Inverse" of the inscribed angle theorem.While doing geometry problem I encountered something I would call "inverse" of inscribed angle theorem. At first I wanted to state it like this:Given △ ABC and point O such that ∠ BOC = 2 ∠ BAC it implies that O is the centre of the circle described on the △ ABCBut of course it is easy to show that this statement doesn't hold. But on my native language forum I found that this theorem should hold:Given △ ABC and ∠ A = α and point O lying on the perpendicular bisector of the segment BC with ∠ BOC = 2 α and point O lying on the same side of line BC as point A then point O is the centre of the circle described on △ ABC

Question

&quot;Inverse&quot; of the inscribed angle theorem.While doing geometry problem I encountered something I would call &quot;inverse&quot; of inscribed angle theorem. At first I wanted to state it like this:Given   △ ABC and point O such that   ∠ BOC = 2   ∠ BAC it implies that O is the centre of the circle described on the   △ ABCBut of course it is easy to show that this statement doesn&#039;t hold. But on my native language forum I found that this theorem should hold:Given   △ ABC and   ∠ A =   α and point O lying on the perpendicular bisector of the segment BC with   ∠ BOC = 2   α and point O lying on the same side of line BC as point A then point O is the centre of the circle described on   △ ABC

Erika Brady · Accepted Answer

Step 1Suppose there is a point       O    ′    ≠  O satisfying the conditions.Step 2Then   C  B      O    ′    =  C  B  O  =  90  −  α so       O    ′    B  O  =  0 and then   O  =      O    ′  .

While doing geometry problem I encountered something I would call "inverse" of inscribed angle theorem.

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