Question

Answer 1

a: Ta có: yy'\(\perp\)MN

zz'\(\perp\)MN

Do đó: yy'//z'z

b: Ta có: zz'//yy'

=>\(\widehat{xAM}=\widehat{ABN}\)(hai góc đồng vị)

=>\(\widehat{ABN}=60^0\)

Ta có: \(\widehat{ABN}+\widehat{ABz}=180^0\)(hai góc kề bù)

=>\(\widehat{ABz}=180^0-60^0=120^0\)

c:

Ta có: zz'//yy'

=>\(\widehat{ABz}=\widehat{MAB}\)(hai góc so le trong)

=>\(\widehat{MAB}=120^0\)

AH là phân giác của góc MAB

=>\(\widehat{HAB}=\widehat{HAM}=\dfrac{\widehat{MAB}}{2}=\dfrac{120^0}{2}=60^0\)

Xét ΔABH có \(\widehat{AHN}\) là góc ngoài tại H

nên \(\widehat{AHN}=\widehat{HAB}+\widehat{HBA}=60^0+60^0=120^0\)