Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta AHD$ có:

Chung $AH$

$\widehat{AHB}=\widehat{AHD}=90^o$ vì $AH\perp BC$

$HB=HD$

$\to\Delta AHB=\Delta AHD(c.g.c)$

$\to AB=AD$

b.Xét $\Delta AHB,\Delta EHD$ có:

$HA=HE$

$\widehat{AHB}=\widehat{EHD}$

$HB=HD$

$\to\Delta AHB=\Delta EHD(c.g.c)$

$\to\widehat{HDE}=\widehat{HBA}$

$\to AB//DE$

c.Từ câu b$\to DE=AB$

Mà $AB=AD\to DE=AD$

Xét $\Delta CHA,\Delta CHE$ có:

Chung $HC$

$\widehat{AHC}=\widehat{CHE}=90^o$

$HA=HE$

$\to\Delta AHC=\Delta EHC(c.g.c)$

$\to \widehat{CAH}=\widehat{CEH}$

$\to\widehat{IAE}=\widehat{AEK}$

Xét $\Delta AEI,\Delta EAK$ có:

$\widehat{IAE}=\widehat{AEK}$

Chung $AE$

$\widehat{AEI}=\widehat{AED}=\widehat{EAD}=\widehat{EAK}$

$\to\Delta AEI=\Delta EAK(g.c.g)$

$\to EI=AK$

Mà $DA=DE$

$\to DI=EI-DE=AK-AD=DK$

d.Ta có $ID=DK, DA=DE$

$\to\Delta DIK, \Delta DAE$ cân tại $D$

$\to \widehat{DIK}=90^o-\dfrac12\widehat{IDK}=90^o-\dfrac12\widehat{ADE}=\widehat{DEA}$

$\to AE//IK\to IK//AH$

Mà $AH\perp BC\to IK\perp BC$