m AOB=2*60°=120°
mas BOD unghi mare =90°+120°=210⇒BOD unghi mic =360-210=150
BOC=mas BOD/3=150/3=50°
(am tinut cont ca mas∡ COD = 2 mas∡ BOC)
mas DOM=90+60=150°=mas ∡BOD
Fie ON semidreapta opusa OD
atunci mas ∡(MON)= 180°-mas ∡AOD- mas∡AOM= 180°-90°-60°=30°
cum mas∡ MOB=60° si 60°2=30°⇒ ON bisectoare
