Xét tứ giác OINJ
có \(\widehat{O}=60^o,\widehat{I}=90^o,\widehat{J}=90^o\)
\(\Rightarrow\widehat{N}=360^o-\left(60^o+90^o+90^o\right)\)
\(=120^o\)
Xét tam giác INJ
có \(\widehat{N}=120^o\Rightarrow\widehat{I_3}+\widehat{J_2}=180^o-\widehat{N}=60^o\)
\(\Rightarrow\widehat{2I_3}+\widehat{2J_2}=120^o\)
Xét tam giác IAR
\(\widehat{IAR}=\widehat{JIA}+\widehat{IJA}\)
\(=\widehat{2I_3}+\widehat{2J_2}=120^o\)
Vậy \(\widehat{IAR}=120^o\)