\(\left\{{}\begin{matrix}AB=BC=CD=DA\\AE=BF=CG=DH\end{matrix}\right.\Rightarrow EB=FC=GD=HA\\ \left\{{}\begin{matrix}AE=BF\\AH=EB\\\widehat{A}=\widehat{B}=90^0\end{matrix}\right.\Rightarrow\Delta AEH=\Delta FBE\\ \Rightarrow EH=EF\)
Cmtt: \(\Delta FBE=\Delta GCF;\Delta GCF=HDG\)
Do đó \(EH=EF=GF=HG\)
Hay EFGH là hình thoi
Mà \(\Delta AEH=\Delta FBE\) nên \(\widehat{AEH}=\widehat{BFE}\)
Mà \(\widehat{BFE}+\widehat{BEF}=90^0\left(\Delta BEF\bot B\right)\)
\(\Rightarrow\widehat{AEH}+\widehat{BEF}=90^0\\ \Rightarrow\widehat{HEF}=180^0-\left(\widehat{AEH}+\widehat{BEF}\right)=90^0\left(\text{kề bù}\right)\)
Vậy EFGH là hình vuông
