In ΔFED dreptunghic in D:
stim de la punctul b) ca EF=ED=> <EDF=<EFD=45°
in ΔAED avem:
<EAD=15°, <EDA=45°=> <AED=180-15-45=120°
in ΔDEB:
<BED=<BEA-<AED=180-120=60°
s-a demonstrat la punctul a) ca BE=FE si la pctb) ca EF=ED=> BE=ED=>
<EDB=<EBD=(180-<DAB):2=(180-60):2=60
=> <EDB=<EBD=<BED=60=> ΔDEB= echilateral