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 \(\widehat{A_2}=90^o\) ) (1)
Trong \(\Delta AKC\) có: \(\widehat{A_3}+\widehat{C_1}=90^o\) ( do \(\widehat{K}=90^o\) ) (2)
Từ (1) và (2) \(\Rightarrow\widehat{A_1}=\widehat{C_1}\)
Xét \(\Delta AHB,\Delta CKA\) có:
\(\widehat{A_1}=\widehat{C_1}\left(cmt\right)\)
AB = AC ( gt )
\(\widehat{H}=\widehat{K}=90^o\)
\(\Rightarrow\Delta AHB=\Delta CKA\) ( c.huyền - g.nhọn )
\(\Rightarrow AH=CK\) ( cạnh t/ứng ) ( đpcm )
b) Vì \(\Delta AHB=\Delta CKA\)
\(\Rightarrow BH=AK,AH=CK\) ( cạnh t/ứng )
Ta có: \(HK=AK+AH=BH+CK\left(đpcm\right)\)
Vậy...
Thanks bạn nhiều
Nguyễn Huy Tú
