Dear Mathlinkers, for (ii) https://artofproblemsolving.com/community/c6t48f6h1491907_an_incenter_problem Sincerely Jean-Louis

Dear Mathlinkers, (i) refers to the Haruki's lemma Sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Quickies%205.pdf p. 20-25 Sincerely Jean-Louis

My solution $(i)$ We know that $\frac{AI}{ID}=\frac{AC}{DC}$ and $EB=EC=EI$, $\triangle DEC \sim \triangle CEA$ It implies that $\frac{AI}{IE}=\frac{ID}{DE}$. $(ii)$ Draw diameter $AJ$, $JF$ cuts $AE$ at $L$. $EF$ cuts $BC$ at $K$. By changing angle we get $\angle I'IL=\angle EFJ \Rightarrow FILI'$ is cyclic On the other hand, $EI^2=EC^2=EK.EF \Rightarrow \angle IKE=\angle FIE=\angle LI'E \Rightarrow KI \parallel LI'$ By Menelaus and Thales, we obtain $\frac{AI}{IE}=\frac{ID}{DE}=\frac{KI'}{KE}=\frac{IL}{IE} \Rightarrow IL=IA=IF$ Q.E.D

nguyendangkhoa17112003 wrote: My solution $(i)$ We know that $\frac{AI}{ID}=\frac{AC}{DC}$ and $EB=EC=EI$, $\triangle DEC \sim \triangle CEA$ It implies that $\frac{AI}{IE}=\frac{ID}{DE}$. $(ii)$ Draw diameter $AJ$, $JF$ cuts $AE$ at $L$. $EF$ cuts $BC$ at $K$. By changing angle we get $\angle I'IL=\angle EFJ \Rightarrow FILI'$ is cyclic On the other hand, $EI^2=EC^2=EK.EF \Rightarrow \angle IKE=\angle FIE=\angle LI'E \Rightarrow KI \parallel LI'$ By Menelaus and Thales, we obtain $\frac{AI}{IE}=\frac{ID}{DE}=\frac{KI'}{KE}=\frac{IL}{IE} \Rightarrow IL=IA=IF$ Q.E.D Nice Mine was a two page bash

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/78.1%20Relations%20metriques.pdf p. 44... Sincerely Jean-Louis

solved also here