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Prove that the points $A, B, C, K$ are concyclic.

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?

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  • $\begingroup$ (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. $\endgroup$ Commented Feb 7, 2022 at 16:03
  • $\begingroup$ The hint that was given in my textbook wasn’t related to my method. @CalvinLin $\endgroup$ Commented Feb 7, 2022 at 18:13
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    $\begingroup$ 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.) $\endgroup$ Commented Feb 7, 2022 at 18:26

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You were heading to a right path, but you drifted. When you added your phantom point $K^\prime$, that is, the intersection of the bisector of $\angle BAC$ and the circumcircle of $\triangle ABC$, you figured out correctly that $BK^\prime=K^\prime C$. You proved it by some angle chasing, tho it still can be proved by the $\textit{Incenter/Excenter Lemma}$ such that $K^\prime$ is the center of the circle containing points $B$, $C$, and $I$ which is the incenter of $\triangle ABC$. Anyways, since $K^\prime$ is equidistant from points $B$ and $C$, then $K^\prime$ has to lie on the perpendicular bisector line of $\overline{BC}$. In other words, it follows that if the intersection of bisector of $\angle BAC$ and perpendicular bisector of $\overline{BC}$ is $K$, then points $A$, $B$, $C$, $K$ are concyclic. $\blacksquare$ enter image description here

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