Dựng tam giác CDG bằng tam giác ADE

Khi đó \(\widehat{ADE}=\widehat{CDG}\); \(DE=DG\); \(\widehat{AED}=\widehat{DGC}=70^o\) hay \(\widehat{DGF}=70^o\)

Dễ thấy \(\widehat{ADE}+\widehat{CDE}=\widehat{ADC}=90^o\) 

\(\Rightarrow\widehat{CDE}+\widehat{CDG}=90^o\Rightarrow\widehat{EDG}=90^o\)

\(\Rightarrow\widehat{GDF}=\widehat{EDG}-\widehat{EDF}=90^o-45^o=45^o\)\(\Rightarrow\widehat{EDF}=\widehat{GDF}\left(=45^o\right)\)

\(\Delta DEF\) và \(\Delta DGF\) có \(DE=DG\) (cmt); \(\widehat{EDF}=\widehat{GDF}\) (cmt) và DF chung

\(\Rightarrow\Delta DEF=\Delta DGF\left(c.g.c\right)\) \(\Rightarrow\widehat{DFE}=\widehat{DFG}\)

Lại có \(\widehat{DFG}=180^o-\left(\widehat{GDF}+\widehat{DGF}\right)=180^o-\left(45^o+70^o\right)=65^o\) \(\Rightarrow\widehat{DFE}=65^o\)

Vậy số đo của góc "?" là 65o.