Răspuns:
Explicație pas cu pas:
ai ∡AOB=1/3×∡BOC=1/2×∡COD=1/6×∡DOE
asta inseamna
∡AOB=∡BOC/3=∡COD/2=∡DOE/6
mai este in ipoteza ∡AOE unghi alungit. =>∡AOE=180°
dar ∡AOE=∡AOB+∡BOC+∡COD+∡DOE
ne uitam atent in relatia ∡AOB=∡BOC/3=∡COD/2=∡DOE/6 si o vom egala cu o constanta k
deci ∡AOB=∡BOC/3=∡COD/2=∡DOE/6=k
asta inseamna ca ∡AOB=K
∡BOC/3=K=>∡BOC=3K
∡COD/2=K=>∡COD=2K
∡DOE/6=K=> ∡DOE=6k
acum avem ∡AOE=∡AOB+∡BOC+∡COD+∡DOE
∡AOE=k+3k+3k+6k
∡AOE=12K
180°=12K=>K=180°:12=> K=15
daca k=15° si ∡AOB=K=>∡AOB=15°
K=15° SI ∡BOC=3K=> ∡BOC=3×15°=> ∡BOC=45°
K=15° SI ∡COD=2K=>∡COD=2×15°=> ∡COD=30°
K=15 SI ∡DOE=6K=> ∡DOE=6×15°=> ∡DOE=90°
∡AOB+∡BOC+∡COD=∡AOD
15°+45°+30°=∡AOD
90°=∡AOD=> AO⊥OD
daca ∡DOE=90°=> DO⊥OE
ca idee cam asa arata figura
