Xét tứ giác OINJ
có \(\widehat{O}=60^o;\widehat{I}=90^0;\widehat{J}=90^o\)
\(\Rightarrow\widehat{N}=360^o-\left(60^o+90^o+90^o\right)\)
\(\Rightarrow\widehat{N}=120^o\)
Xét \(\Delta NIJ\) \(\widehat{N}=120^o\Rightarrow\widehat{I3}+\widehat{J2}=180^o-\widehat{N}=60^o\)
\(\Rightarrow\widehat{2I3}+\widehat{2J2}=120^o\)
Xét \(\Delta IAR\)
\(\widehat{IAR}=\widehat{JIA}+\widehat{IJA}=\widehat{2I3}+\widehat{2J2}=120^o\)
Vậy \(\widehat{JAR}=120^o\)
Xét tứ giác OINJ
có ˆO=60o;ˆI=900;ˆJ=90oO^=60o;I^=900;J^=90o
⇒ˆN=360o−(60o+90o+90o)⇒N^=360o−(60o+90o+90o)
⇒ˆN=120o⇒N^=120o
Xét ΔNIJΔNIJ ˆN=120o⇒ˆI3+ˆJ2=180o−ˆN=60oN^=120o⇒I3^+J2^=180o−N^=60o
⇒ˆ2I3+ˆ2J2=120o⇒2I3^+2J2^=120o
Xét ΔIARΔIAR
ˆIAR=ˆJIA+ˆIJA=ˆ2I3+ˆ2J2=120oIAR^=JIA^+IJA^=2I3^+2J2^=120o
Vậy ˆJAR=120o
