Giải thích các bước giải:

a.Xét $\Delta ABH,\Delta EBH$ có:

Chung $BH$

$BA=BE$

$HA=HE$ vì $H$ là trung điểm $AE$
$\to \Delta ABH=\Delta EBH(c.c.c)$

b.Từ câu a $\to \widehat{AHB}=\widehat{BHE}$

Mà $\widehat{AHB}+\widehat{BHE}=180^o$

$\to \widehat{AHB}=\widehat{BHE}=90^o$

$\to BH\perp AE$

c.Từ câu a $\to \widehat{ABH}=\widehat{EBH}\to \widehat{ABD}=\widehat{EBD}$

Xét $\Delta ABD,\Delta EBD$ có:

Chung $DB$

$\widehat{ABD}=\widehat{EBD}$

$BA=BE$

$\to \Delta ABD=\Delta EBD(c.g.c)$

$\to\widehat{BED}=\widehat{BAD}=90^o$

$\to DE\perp BC$

$\to\widehat{DEC}=90^o$

d.Từ câu c $\to DA=DE$

Xét $\Delta DAK,\Delta DEC$ có:

$DA=DE$

$\widehat{ADK}=\widehat{EDC}$

$DK=DC$

$\to \Delta DAK=\Delta DEC(c.g.c)$

$\to\widehat{DAK}=\widehat{DEC}=90^o$

$\to \widehat{BAK}=\widehat{BAD}+\widehat{DAK}=90^o+90^o=180^o$

$\to B, A, K$ thẳng hàng