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The incircle of $\triangle ABC$ with incentre $I$ touches the sides $BC,CA$ and $AB$ at $D,E$ and $F$ respectively. Now, let $K=BI\cap EF$. Show that $BK\perp CK$.

Hint: Bonus problem: Let the reflection of $C$ through $K$ be $C_1$. Prove that $A,B$ and $C_1$ are collinear.

Let $EF \cap BC=L$.Then $(C,B;D,L)=-1$ and $\angle LKB=\angle BKD$,So,$BK \perp KC$.

Thanic Nur Samin wrote: Bonus problem: Let the reflection of $C$ through $K$ be $C_1$. Prove that $A,B$ and $C_1$ are collinear.

Let $CK \cap AB=C_{1}$.$\Delta C_{1}BK$ is the reflection of $\Delta CBK$ across $BK$.So,the reflection of $C$ through $K$ is $C_1$.