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In scalene triangle $ABC$, let $K$ be the intersection of the angle bisector of $∠A$ and the perpendicular bisector of $BC$. Prove that the points $A, B, C, K$ are concyclic.

$D$ is the midpoint of $[BC]$. Let $K’$ be the intersection of the angle bisector of $∠A$ with $(ABC)$, we’ll show that $(K’O) \perp (BC)$.

$ABK’C$ is cyclic $\implies \angle K’AC=K’BC=a$ but $\angle BCK’=a$, hence $$BK’=K’C\implies \triangle BDK’\sim \triangle K’DC\implies \angle BDK’ = \angle K’DC$$ Since $\angle BDK ’+\angle K’DC=180^{o}$, we get $\angle K’DC=90^{o}$. Is my proof correct?

PNT
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  • (I've not read through your proof, but) can you explain what concerns you have about whether your proof is correct or wrong? Is there a step that you think is suspicious? Are you unsure about applying a certain technique? Are you concerned you might be missing a case? If so, please point it out. – Calvin Lin Feb 07 '22 at 16:03
  • The hint that was given in my textbook wasn’t related to my method. @CalvinLin – PNT Feb 07 '22 at 18:13
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    Then you should make that clear. Of course, we don't always need to follow the hint to solve a problem. This is a "well-known" olympiad fact that has several approaches. $\quad$ Your approach works. You should make it clear that $K', K$ coincide. (IE You haven't shown that ABCK are concyclic.) – Calvin Lin Feb 07 '22 at 18:26

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