Đáp án:
a)
Ta có:
$\widehat{ABC}+\widehat{ABD}=180^0$
$\widehat{ACB}+\widehat{ACE}=180^0$
mà $\widehat{ABC}=\widehat{ACE}$
$\Rightarrow \widehat{ABD}=\widehat{ACE}$
b)
Xét $\triangle ABD$ và $\triangle ACE$ có
$AB=AC$
$\widehat{ABD}=\widehat{ACE}$
$BD=CE$
$\Rightarrow \triangle ABD=\triangle ACE$ (c.g.c)
c)
Do $\triangle ABD=\triangle ACE$ (cmt)
$\Rightarrow \widehat{ADB}=\widehat{AEC}$
$\Rightarrow \triangle ADE$ cân tại $A$
d)
Xét $\triangle BDH$ và $CEK$ có
$\widehat{BHD}=\widehat{CKE}=90^0$
$BD=CE$
$\widehat{HDB}=\widehat{CEK}$ (do $ \widehat{ADB}=\widehat{AEC}$-cmt)
$\Rightarrow \triangle BDH=\triangle CEK$ (cạnh huyền - góc nhọn)
$\Rightarrow BH=CK$