Question

dau 100 puncte si coroana cine rezolva corect exercitiul 8

Answer 1

∡AOB si ∡BOC complementare ⇒m(∡AOB)+m(∡BOC)=90°
(OD bisect ∡AOB ⇒m(∡DOB)=m(∡AOB)/2 (1)
(OF bisect ∡BOC ⇒m(∡BOF)=m(∡BOC)/2 (2)
Adunand (1) si (2) obtinem
m(∡DOB)+m(∡BOF)=m(∡DOF)=(m(∡AOB)+m(∡BOC))/2=90°/2=45°

Answer 2

m(∡AOB)+m(∡BOC)=90°
m(∡AOD)=m(∡DOB)=m(∡AOB):2
m(∡BOF)=m(∡FOC)=m(∡BOC):2
m(∡DOF)=m(∡DOB)+m(∡BOF)=m(∡AOB):2+m(∡BOC):2=
=[m(∡AOB)+m(∡BOC)]:2=90°:2=45°