Pb 4. Figura e un dreptunghi ABCD, daca < ABD=60 grd, atunci <ADB=30 grd
trg ABD=trg ABC :{ AB comuna si diag AC si BD congr
rezulta < ACB=30 grd si <CAB=60 grade
in trg AMB, <ABM=30 grade=90-60 adica BM e bisectoarea < ABD
5. <AOB=180= 3x (<AOC=2x, <COB=x)
rezulta x=60
<BOD=<AOC (op la varf
)=2*60=120 grd
OF bisectoare rezulta ca <BOF=60 grd
dar stiu ca <BOE=90 grd, rezulta < EOF=EOB-FOB=90-60=30
9. notez <AOB=2x
<BOD=180-2x
OE bisectoarea <BOD, rezulta < DOE=<EOB=(180-2x)/2= 90-x
<FOE=90-x-30=60-x
<COF=<FOE (OF bisect COE)
COF=60-x
2x+30=60-x
3x=30, x=10, <AOB=2*10=20 grd
daca OH bisect COD, rezulta ca <HOD=10 grade
dar DOE=80 grd
<HOE=<HOD+<DOE=10+80=90, rezulta ca OH_|_ OE
