Let O=circumcentre of $\triangle{ABC}$ DO=perpendicular bisector of(BC) Let K'=DO$\cap$BI By a simple angle chasing we get $\angle$BK'D=$\angle$ICD so IDCK' is cyclic =>$\odot$CDI$\cap$BI={I,K'}=>K=K'=>BK=CK

$DIKC$ is cyclic: $\widehat{KCD}=\widehat{BID}$ But, $K\widehat CD = K\widehat CB + B\widehat CD = K\widehat CB + \frac{{\widehat A}}{2}$ and $B\widehat ID = \frac{{arc(AE)}}{2} + \frac{{arc(BD)}}{2} = \frac{{\widehat B}}{2} + \frac{{\widehat A}}{2} = K\widehat BC + \frac{{\widehat A}}{2}$ Hence: $K\widehat CB = K\widehat BC \Leftrightarrow $ $\boxed{BK=CK}$

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george_54 wrote: $DIKC$ is cyclic: $\widehat{KCD}=\widehat{BID}$ But, $K\widehat CD = K\widehat CB + B\widehat CD = K\widehat CB + \frac{{\widehat A}}{2}$ and $B\widehat ID = \frac{{arc(AE)}}{2} + \frac{{arc(BD)}}{2} = \frac{{\widehat B}}{2} + \frac{{\widehat A}}{2} = K\widehat BC + \frac{{\widehat A}}{2}$ Hence: $K\widehat CB = K\widehat BC \Leftrightarrow $ $\boxed{BK=CK}$ This solution is rather more simple than in the source

Easy to se that the triangles KBD and KCD are congruent, then the conclusion follows.

By Incenter/Excenter Lemma, $D$ is the circumcenter of $\triangle IBD \Longrightarrow DI = DC$ $KICD$ is a cyclic quadrilateral $\Longrightarrow \angle DIC = \angle DCI = \angle DKC = \angle DKB$ $ACDB$ is a cyclic quadrilateral $\Longrightarrow \angle ADC = \angle ABC = 180 - 2\angle DKB$ $\angle KBC = 90 - \angle DKB \Longrightarrow KD \perp BC$ Hence, $KB = KC$

Dear Mathlinkers, why not with my favorite theorem ''Reim''... Sincerely Jean-Louis

My Solution by Reim's: Let $CK\cap{(ABC)}=C_1$ and let $\angle A=\alpha$ , $\angle B=\beta$ and $\angle C=\gamma$ Then by Reim's we obtain : $CA_1\parallel{BI}$ ,hence $\angle C_1BE=180-\angle BC_1A=\angle BCA=\gamma$ Obviously $\angle BC_1C=\alpha\rightarrow{\angle C_1KB=\angle CKE=\beta}$ Finally consider $\triangle EKC$ , clearly $\angle KEC=\alpha$ , $\angle CKE=\beta$ , hence $\angle KCE=\gamma=\angle BCA$ So $\angle KCB=\angle ACE=\angle ABE=\angle EBC\equiv{\angle KBC}$ , as desired . $\blacksquare$

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Dear, with the Reil's theorem we don't need angles... Sincerely Jean-Louis

jayme wrote: why not with my favorite theorem ''Reim''... Let $BI, CK$ meet $(ABC)$ at $E,F$ resp. By our favorite Reim $AF \parallel IK$. So $\angle CBK = \angle EBA = \angle FCB$. Remark : $I$ can be any point on bisector of $\angle B$.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/26.%203.%20Segments.pdf p. 6... Sincerely Jean-Louis

Just simple angle chasing. The diagram given by george_54 may be followed for notations to this answer. As $A,B,C\ $ and $\ D$ lie on $ \odot ABC $, we can say that $ \angle CDA = \angle CBA =\angle B $. Now, as $C,D,I$ and $K$ are concyclic, we have $\angle BKC = \angle IKC = 180^\circ -\angle CDI = 180^\circ -\angle CDA = -\angle B $ In $\triangle BKC,$ $ \angle CBK + \angle BKC + \angle KCB =180^\circ $ $\Rightarrow \angle \frac{B}{2} - \angle B + \angle KCB =180^\circ $ $\Rightarrow \angle KCB = \angle \frac{B}{2} = \angle CBK $ Therefore, $BK = CK $.

My Solution: Let $\angle BAC=2\alpha$ and $\angle ABC=2\beta$. Using Incenter-Excenter Lemma,$\angle IBD=\alpha+\beta=\angle BID\Rightarrow \angle KID=180^\circ-\alpha-\beta$ Let $\angle IKD=\gamma\Rightarrow \angle IDK=180^\circ-(180^\circ-\alpha-\beta)-\gamma=\alpha+\beta-\gamma$ $\Rightarrow \angle IDK=\angle ICK=\alpha+\beta-\gamma\Rightarrow \angle KCB=\angle KCI+\angle ICB=\alpha+\beta-\gamma+90^\circ-\alpha-\beta=90^\circ-\gamma\Rightarrow \angle KCB=90^\circ-\gamma$. Note,that $\gamma=\angle IKD=\angle ICD=90^\circ-\beta\Rightarrow \angle KCB=90^\circ-(90^\circ-\beta)=\beta\Rightarrow BK=KC$ $Q.E.D$

Dear Mathlinkers, https://artofproblemsolving.com/community/c4t48f4h1845476_geome p. 5-6. Sincerely Jean-Louis

∠BID = ∠IBA + ∠IAB = ∠KCD ∠BCD = ∠IAB ---> ∠KCB = ∠IBA = ∠IBC ---> KCB is isosceles ---> KB = KC.