Giải:
a, Ta có: \(\widehat{KBC}+\widehat{KCB}=90^o\left(\widehat{BKC}=90^o\right)\)
\(\widehat{KBC}+\widehat{ABH}=90^o\)
\(\Rightarrow\widehat{KCB}=\widehat{ABH}\)
Tương tự \(\widehat{HAB}=\widehat{KBC}\)
Xét \(\Delta HAB,\Delta KBC\) có:
\(\widehat{KCB}=\widehat{ABH}\left(cmt\right)\)
AB = AC ( gt )
\(\widehat{HAB}=\widehat{KBC}\left(cmt\right)\)
\(\Rightarrow\Delta HAB=\Delta KBC\left(g-c-g\right)\)
\(\Rightarrow BH=CK\left(đpcm\right)\)
b, Gọi giao giữa AM, BH là I
t/g cân ABC tại M có BM là phân giác
\(\Rightarrow\)BM là đường cao
\(\Rightarrow\widehat{IBM}+\widehat{BIM}=90^o\left(\widehat{IMB}=90^o\right)\)
\(\widehat{HAI}+\widehat{AIH}=90^o\left(\widehat{AHI}=90^o\right)\)
Mà \(\widehat{AIH}=\widehat{BIM}\) ( đối đỉnh )
\(\Rightarrow\widehat{IBM}=\widehat{HAI}\) hay \(\widehat{KBM}=\widehat{HAM}\)
Xét \(\Delta AHM,\Delta BKM\) có:
AH = BK ( theo a )
\(\widehat{KBM}=\widehat{HAM}\left(cmt\right)\)
\(AM=BM\left(=\dfrac{1}{2}BC\right)\)
\(\Rightarrow\Delta AHM=\Delta BKM\left(c-g-c\right)\)
\(\Rightarrow HM=KM\) (1)
và \(\widehat{AMH}=\widehat{BMK}\)
Mà \(\widehat{BMK}+\widehat{KMI}=90^o\)
\(\Rightarrow\widehat{AMH}+\widehat{KMI}=90^o\)
\(\Rightarrow\widehat{HMK}=90^o\) (2)
Từ (1), (2) \(\Rightarrow\Delta HKM\) vuông cân tại M ( đpcm )
Vậy...
