Giải thích các bước giải:

a.Xét $\Delta ABD,\Delta HDB$ có:

$\widehat{BAD}=\widehat{BHD}(=90^o)$

Chung $BD$

$\widehat{ABD}=\widehat{HBD}$

$\to \Delta ABD=\Delta HBD$(cạnh huyền-góc nhọn)

$\to BA=BH, DA=DH$

$\to B, D\in$ trung trực $AH$

$\to BD$ là trung trực $AH$

$\to BD\perp AH$

c.Xét $\Delta ACE, \Delta KCE$ có:

$\widehat{ACE}=\widehat{KCE}$

Chung $CE$

$\hat A=\hat K(=90^o)$

$\to \Delta ACE=\Delta KCE$(cạnh huyền-góc nhọn)

$\to CA=CK, EA=EK$

$\to AB+AC=BH+CK=BH+(HC+HK)=(BH+HC)+HK=BC+HK$

c.Vì $BA=BH, CA=CK$

$\to \Delta BAH,\Delta CAK$ cân tại $B,C$

$\to \widehat{BAH}=90^o-\dfrac12\hat B$

     $\widehat{CAK}=90^o-\dfrac12\hat C$

$\to \widehat{BAH}+\widehat{CAH}=180^o-\dfrac12(\hat B+\hat C)$

$\to \widehat{BAH}+\widehat{HAC}+\widehat{HAK}=180^o-\dfrac12\cdot 90^o$

$\to \widehat{BAC}+\widehat{HAK}=180^o-45^o$

$\to 90^o+\widehat{HAK}=135^o$

$\to \widehat{HAK}=45^o$