)
Giải:
Gọi giao điểm giữa KC và AB là O
giao điểm giữa KC và BD là I
a) Ta có: \(\widehat{KAC}=\widehat{A_1}+\widehat{A_2}=90^o+\widehat{A_2}\)
\(\widehat{DAB}=\widehat{A_3}+\widehat{A_2}=90^o+\widehat{A_2}\)
\(\Rightarrow\widehat{KAC}=\widehat{DAB}\)
Xét \(\Delta ACK,\Delta ABD\) có:
\(AK=AB\left(gt\right)\)
\(\widehat{KAC}=\widehat{DAB}\left(cmt\right)\)
\(AD=AC\left(gt\right)\)
\(\Rightarrow\Delta ACK=\Delta ABD\left(c-g-c\right)\)
b) Vì \(\Delta ACK=\Delta ABD\)
\(\Rightarrow\widehat{K_1}=\widehat{B_1}\) ( góc t/ứng )
Xét \(\Delta KAO\) có: \(\widehat{D_1}+\widehat{A_1}+\widehat{O_1}=180^o\)
Xét \(\Delta BOI\) có: \(\widehat{B_1}+\widehat{I_1}+\widehat{O_2}=180^o\)
Mà \(\widehat{B_1}=\widehat{D_1}\left(cmt\right);\widehat{O_1}=\widehat{O_2}\) ( đối đỉnh )
\(\Rightarrow\widehat{A_1}=\widehat{I_1}\)
Mà \(\widehat{A_1}=90^o\Rightarrow\widehat{I_1}=90^o\)
\(\Rightarrow KI\perp BI\) hay \(KC\perp BD\left(đpcm\right)\)
Vậy...
