Giải:
a) Xét \(\Delta BAC,\Delta ECA\) có:
\(AB=CE\left(gt\right)\)
\(\widehat{BAC}=\widehat{ECA}\left(=90^o\right)\)
\(AC\): cạnh chung
\(\Rightarrow\Delta BAC=\Delta ECA\left(c-g-c\right)\)
\(\Rightarrow BC=AE\) ( cạnh t/ứng ) ( đpcm )
\(\Rightarrow\widehat{BCA}=\widehat{EAC}\) ( góc t/ứng )
Mà 2 góc trên ở vị trí so le trong nên BC // AE ( đpcm )
b) Ta có: \(\widehat{EAC}+\widehat{ECA}=\widehat{AEx}\) ( góc ngoài \(\Delta ECA\) )
\(\Rightarrow\widehat{EAC}+90^o=120^o\)
\(\Rightarrow\widehat{EAC}=30^o\)
Mà \(\widehat{BCA}=\widehat{EAC}\Rightarrow\widehat{BCA}=30^o\)
Xét \(\Delta ABC\) có: \(\widehat{BCA}+\widehat{ABC}=90^o\) ( do \(\widehat{A}=90^o\) )
\(\Rightarrow\widehat{ABC}=60^o\) ( do \(\widehat{BCA}=30^o\) )
Vậy...
