$a)$Xét $\Delta AEB$ và $\Delta DEC$
$AE=DE\\ EB=EC\\ \widehat{E_1}=\widehat{E_2}(đđ)\\ \Rightarrow \Delta AEB = \Delta DEC\\ b)\Delta AEB = \Delta DEC\\ \Rightarrow \widehat{A_1}=\widehat{D_1}\\ \Rightarrow AB//CD\\ c)AB//CD\\ AB \perp AC\\ \Rightarrow CD \perp AC\\ \Rightarrow \widehat{DCA}=90^o\\ \Delta AEB = \Delta DEC\\ \Rightarrow AB=DC$
Xét $\Delta BAC$ và $\Delta DCA$
AC: chung
$\widehat{BCA}=\widehat{DCA}=90^o\\ AB=DC\\ \Rightarrow \Delta BAC = \Delta DCA\\ d)\Delta BAC = \Delta DCA\\ \Rightarrow BC=AD\\ \Rightarrow BE=EC=\dfrac{BC}{2}=\dfrac{AD}{2}=AE=ED$
$\Rightarrow \Delta AEC, \Delta BEC$ cân tại $E$
$\Rightarrow \widehat{A_2}=\dfrac{180^o-\widehat{E_3}}{2};\widehat{D_2}=\dfrac{180^o-\widehat{E_4}}{2}$
Mà $\widehat{E_3}=\widehat{E_4}(đđ)$
$\Rightarrow \widehat{A_2}=\widehat{D_2}$
$\Rightarrow AC//BD$
Mà $CD \perp AC$
$\Rightarrow CD \perp BD$
$\Rightarrow \Delta BDC$ vuông tại $D$
