Ta có hình vẽ:
Ta có: Vì \(EH\perp BC\) nên \(\widehat{BHE}=90^o\)
nên \(\widehat{A}=\widehat{BHE}=90^o\)
và \(\widehat{ABE}=\widehat{HBE}=\dfrac{1}{2}\widehat{B}\)(gt)
Ta có: \(\left\{{}\begin{matrix}\widehat{BEH}=180^o-\widehat{HBE}-\widehat{BHE}\\\widehat{BEA}=180^o-\widehat{A}-\widehat{ABE}\end{matrix}\right.\)
Suy ra: \(\widehat{BEH}=\widehat{BEA}\)
Xét 2 tam giác \(BEA\) và \(BEH\) ta có:
\(\left\{{}\begin{matrix}BE-chung\\\widehat{BEA}=\widehat{BEH}\left(cmt\right)\\\widehat{ABE}=\widehat{HBE}\left(gt\right)\end{matrix}\right.\Leftrightarrow\Delta BEA=\Delta BEH\left(g.c.g\right)\)
Suy ra \(AB=BH\)(2 cạnh tương ứng)
b) Ta có: \(\widehat{BEA}=\widehat{BEH}\left(cmt\right)\) mà \(\widehat{BEA}+\widehat{BEH}=\widehat{AEH}=120^o\Leftrightarrow\widehat{BEA}=\widehat{BEH}=60^o\)
\(\widehat{ABE}=180^o-\widehat{A}-\widehat{BEA}=180^o-90^o-60^o=30^o\)
Xét \(\Delta BAE\) và \(\Delta BHE\) có :
BE : cạnh chung
\(\widehat{ABE}=\widehat{HBE}\) (gt)
Ta có : \(\widehat{ABE}+\widehat{BAE}+\widehat{AEB}=\widehat{HBE}+\widehat{BHE}+\widehat{HEB}\)
\(\Rightarrow\widehat{BEA}=\widehat{BEH}\)
\(\Rightarrow\Delta BAE=\Delta BHE\) (g . c . g)
\(\Rightarrow\) AB = BH
Vì \(\Delta BAE\) và \(\Delta BHE\) là tam giác vuông .
\(\Rightarrow\widehat{ABE}+\widehat{AEB}=\widehat{EBH}+\widehat{HEB}\)
\(\Rightarrow\widehat{AEB}=\widehat{HEB}=120^0\)
\(\Rightarrow\widehat{AEB}=\)\(\dfrac{1}{2}\)\(\times120^0=60^0\)
Mà \(\widehat{ABE}+\widehat{BAE}+\widehat{AEB}=180^0\)
\(\Rightarrow\widehat{ABE}=180^0-90^0-60^0=30^0\)
