Question

Giải giúp mình ạ

 

Answer 1

a) Ta có: Ax⊥MN,Cy⊥MN

=> Ax//Cy

b) Ta có: Bz//Ax

\(\Rightarrow\widehat{ABz}=\widehat{BAx}=60^0\)(so le trong)

c) Ta có: Bz//Ax,Ax//Cy

=> Bz//Cy

\(\Rightarrow\widehat{CBz}=\widehat{BCy}=30^0\)(so le trong)

\(\Rightarrow\widehat{ABC}=\widehat{ABz}+\widehat{CBz}=60^0+30^0=90^0\)

=> BA⊥BC