<Tri-2023-2038>, hint

Suppose lines AJ and AK cut circle in J' and K" respectively.

<AC'C=90-A/6, <AB'B=<ABB'=90-A/3 and so <KC'C=180-A/2. A,K, C',C are concyclic 

<CKC' =A/3. <CBJ'=A/3. KC'//BB'. Hence B'=J'


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