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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?

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?

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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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$$\angle BDK ’+\angle K’DC=180^{o}$, we get $\angle K’DC=90$$\angle K’DC=90^{o}$. Is my proof correct?

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 $\angle BDK’+\angle K’DC=180$, we get $\angle K’DC=90$. Is my proof correct?

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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PNT
PNT
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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 $\angle BDK’+\angle K’DC=180$, we get $\angle K’DC=90$. Is my proof correct?