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I'm trying to solve this problem, I think I almost made it, at least I hope. I don't know where I'm wrong.
Let I be the incenter of a triangle $ABC$ ($AB < AC$). The line $AI$ intersects the circumcircle of $ABC$ again at $D$. The circumcircle of $CDI$ intersects $BI$ again at $K$. Prove that $BK = CK$."

We note that the quadrilateral $DXKC$ is cyclic, so $\angle CKI=\angle CKI$, $\angle CID=\angle KCD$. The point of tangency of the circle with the side $BC$ is $F$. We thus obtain the chord subtended ($FL$) at $\angle FDL$. Now we extend $L$ making it coincide with the center until it intersects the circumference, we call $S$ the point. We obtain that $\angle SFD=\angle SLD$ are therefore equal. Also $\angle CFL=\angle FDL$ (for tangency), $BFS=FLS$ (for tangency). So we have that $\angle FBK=\angle CFL$, but so I have $BC=CK$.

I really don't know where I'm wrong and i dont watch the solution

Math
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1 Answers1

2

enter image description here

In the figure, all blue dotted angles are equal.

$red = 2 \times blues$ (at D)

$red = blue + green$

result follows.

Mick
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