a) In ΔADR , avem < DAR=22°30', <ADR=90°=>
<DRA=180°-<DAR-<ADR=180°-22°30' - 90°=67°30'
in ΔDSR, avem < DRS=67°30'si stim ca < DR=45°=>
<DSR=180°-<DRS- <SDR=180°-67°30'-45°=67°30'
=> <DSR≡<DRS=67°30'
=> ΔDRS= isoscel
b)comparam ΔDAS~ΔDCS
DS=DS=latura comuna
DA≡DC=latura patrat
<ADS≡<CDS=45°=> L.U.L=> ΔDAS≡ΔDCS=> AS≡SC
c)<ASB≡<DSR= 67°30' =unghiuri opse la varf
din ΔSBA ~ΔSDR ( putem calcula <SAB=67°30', dar si prin DR||AB )
=> ΔSBA =isoscel => BS≡BA≡AD=latura patrat (1)
stim e la pct a) ca ΔDSR =isoscel cu DS≡DR (2)
deci din (1) si (2) avem: BD=BS+SD=AD+DR