Giải:
a) Ta có: \(\widehat{A_1}+\widehat{A_2}+\widehat{A_3}=180^o\) ( góc bẹt )
\(\Rightarrow\widehat{A_1}+\widehat{A_3}=90^o\) ( do \(\widehat{A_2}=90^o\) )
Mà \(\widehat{C_1}+\widehat{A_3}=90^o\) ( do t/g CAK có \(\widehat{K}=90^o\) )
\(\Rightarrow\widehat{A_1}=\widehat{C_1}\)
Xét \(\Delta BHA,\Delta AKC\) có:
\(\widehat{H}=\widehat{K}=90^o\)
AB = AC ( gt )
\(\widehat{A_1}=\widehat{C_1}\left(cmt\right)\)
\(\Rightarrow\Delta BHA=\Delta AKC\) ( c.huyền - g.nhọn )
\(\Rightarrow AH=CK\) ( cạnh t/ứng ) ( đpcm )
b) Vì \(\Delta BHA=\Delta AKC\)
\(\Rightarrow BH=AK,CK=AH\) ( cạnh t/ứng )
\(\Rightarrow AK+AH=BH+CK=HK\) ( đpcm )
Vậy...