see attachement.

2. 1) $AB=AD, \ AE=DF, \ \widehat{BAF}=\widehat{ADF}\Rightarrow$

$\Rightarrow\Delta ABE\equiv\Delta DAF\Rightarrow\widehat{AEB}=\widehat{AFD}, \ \widehat{ABE}=\widehat{DAF}$

$\widehat{BPF}=\widehat{APE}=180^{\circ}-(\widehat{EAP}+\widehat{AEP})=180^{\circ}-(\widehat{DAF}+\widehat{AFD})=$

$=\widehat{D}=180^{\circ}-\widehat{C}=120^{\circ}$

2) $\widehat{ADB}=\widehat{ABD}=\widehat{BAC}=\widehat {BCA}=\alpha$

In $\Delta ABD$ we have $3\alpha+90^{\circ}=180^{\circ}\Rightarrow\alpha=30 ^{\circ}$

$\Rightarrow\widehat{DAB}=120^{\circ}\Rightarrow\wi dehat{BCD}=60^{\circ}$