RagvaloD 10.07.2017 16:03 $CL$ is bisector of $\angle C$ of $ABC$ and intersect circumcircle at $K$. $I$ - incenter of $ABC$. $IL=LK$. Prove, that $CI=IK$ D. Shiryaev
Davrbek 10.07.2017 18:25 $\triangle AKL$ similar to $\triangle CKA$. Then we have $\frac{CK}{AK}=\frac{AK}{KL}$.It is easy to see $AK=KI$.From $IL=LK \implies AK=2KL$.Then $\frac{CK}{AK}=\frac{AK}{KL}=2 \implies CK=2AK=2KI \implies CI=KI$.Proved.
jayme 22.05.2018 09:30 Dear Mathlinkers, the kern of this problem is : ABC verify the relation CA + CB = 2.AB... and we have a lot of consequences... Sincerely Jean-Louis
Mahdi_Mashayekhi 02.01.2022 15:45 ∠KAI = ∠KAB + ∠BAI = ∠ACI + ∠IAC = ∠AIK ---> AK = KI. we have AKL and CAK are similar so AK^2 = KL.KC. now we have AK = 2KL so KC = 4KL and CI = KI.