a, Vì BE là tia phân giác \(\widehat{ABC}\Rightarrow\widehat{ABE}=\widehat{CBE}=30^0\Leftrightarrow\widehat{ABE}=\widehat{HBE}\)
Xét \(\Delta ABE\) và \(\Delta HBE\) có:
\(\widehat{ABE}=\widehat{HBE}\) (cmt)
BE là cạnh chung
\(\widehat{BAE}=\widehat{BHE}=90^0\)
\(\Rightarrow\Delta ABE=\Delta HBE\) (cạnh huyền - góc nhọn)
b,
Xét \(\Delta BHE\) có: \(\widehat{BHE}+\widehat{BEH}+\widehat{HBE}=180^0\)
\(\widehat{BEH}=180^0-\widehat{BHE}-\widehat{HBE}=180^0-90^0-30^0=60^0\)
Xét \(\Delta CHE\) có: \(\widehat{CHE}+\widehat{CEH}+\widehat{HCE}=180^0\)
\(\widehat{CEH}=180^0-\widehat{CHE}-\widehat{HCE}=180^0-90^0-30^0=60^0\)
\(\Rightarrow\widehat{BEH}=\widehat{CEH}\)
Xét \(\Delta HBE\) và \(\Delta HCE\) có:
\(\widehat{BEH}=\widehat{CEH}\) (cmt)
HE là cạnh chung
\(\widehat{BHE}=\widehat{CHE}=90^0\)
\(\Rightarrow\Delta HBE=\Delta HCE\) (g.c.g)
\(\Rightarrow HB=HC\) (2 cạnh tương ứng)
\(\)
a) Có: `△ABE=△HBE (ch-gn)`
\(\left\{{}\begin{matrix}BEchung\\\widehat{ABE}=\widehat{HBE}\\\widehat{A}=\widehat{EHB}=90^0\end{matrix}\right.\)
b) Vì `△ABE=△HBE (ch-gn)`
`=> BA=BH(...)`
mà \(\widehat{B}=60^0\)
`=> △ABH` đều
`=>` \(\widehat{BAH}=60^0,BH=AH\)
Có: \(\widehat{BAH}+\widehat{HAE}=\widehat{A}\)
\(\Rightarrow60^0+\widehat{HAE}=90^0\)
\(\Rightarrow\widehat{HAE}=30^0\)
Trong `△ ABC` có: \(\widehat{C}=180^0-\widehat{A}-\widehat{B}=180^0-90^0-60^0=30^0=\widehat{HAE}\)
`=> △AHC` cân
\(\Rightarrow AH=HC=BH\left(đpcm\right)\)
