Giải thích các bước giải:

a.Ta có $AH\perp BC\to\Delta AHB$ vuông tại $H$

$\to \widehat{HAB}=90^o-\hat B=30^o$

b.Xét $\Delta AHI,\Delta ADI$ có:

Chung $AI$

$IH=ID$ vì $I$ là trung điểm $DH$

$AH=AD$

$\to\Delta AHI=\Delta ADI(c.c.c)$

$\to \widehat{AIH}=\widehat{AID}$

Mà $\widehat{AIH}+\widehat{AID}=180^o$

$\to \widehat{AIH}=\widehat{AID}=90^o$

c.Từ câu b $\to\widehat{IAH}=\widehat{IAD}\to\widehat{KAH}=\widehat{KAD}$

Xét $\Delta AHK,\Delta ADK$ có:
Chung $KA$

$\widehat{AKH}=\widehat{KAD}$

$AH=AD$

$\to\Delta AHK=\Delta ADK(c.g.c)$

$\to \widehat{KDA}=\widehat{AHK}=90^o$

$\to DK\perp AC$

Mà $AB\perp AC\to AB//KD$

d.Ta có:

$\widehat{HAD}=90^o-\widehat{HAB}=60^o$

Ta có $\widehat{IAH}=\widehat{IAD}$

$\to \widehat{IAH}=\widehat{IAD}=\dfrac12\widehat{HAD}=30^o$

$\to\widehat{HAI}=\widehat{HAB}$

$\to \widehat{BAH}=\widehat{HAK}$

Xét $\Delta AHB,\Delta HAK$ có:

$\widehat{HAB}=\widehat{HAK}$

Chung $AH$
$\widehat{AHB}=\widehat{AHK}(=90^o)$

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

$\to HB=HK$

Xét $\Delta HAB,\Delta HEK$ có:

$HA=HE$

$\widehat{AHB}=\widehat{KHE}$

$HB=HK$

$\to\Delta HAB=\Delta HEK(c.g.c)$

$\to \widehat{HAB}=\widehat{HEK}$

$\to AB//KE$
Lại có $KD//AB$

$\to D, K, E$ thẳng hàng