Cho Hình 3.24.

a) Ta có $yy'\perp MN$ và $zz'\perp MN$ nên yy'//zz'.

b) Ta có yy' // zz' suy ra $\widehat{xAM}=\widehat{ABN}=60^{\circ}$ (hai góc đồng vị).

Mà $\widehat{ABz}+\widehat{ABN}=180^{\circ}$ (hai góc kề bù) hay $\widehat{ABz}+60^{\circ}=180^{\circ}$

Do đó $\widehat{ABz}=180^{\circ}-60^{\circ}=120^{\circ}$.

c) Ta có yy' // zz' suy ra $\widehat{ABz}=\widehat{BAM}$ (hai góc so le trong) nên $\widehat{BAM}=120^{\circ}$.

Do: AH là phân giác của $\widehat{BAM}$ suy ra $\widehat{HAM}=\widehat{HAB}=\frac{120^{\circ}}{2}=60^{\circ}$

Ta có: yy' //zz' suy ra $\widehat{HAM}=\widehat{AHB}$ (hai góc so le trong) nên $\widehat{AHB}=60^{\circ}$

Ta có: $\widehat{AHB}+\widehat{AHN}=180^{\circ}$ (kề bù)

suy ra $60^{\circ}+\widehat{AHN}=180^{\circ}$ => $\widehat{AHN}=120^{\circ}$

Vậy $\widehat{AHN}=120^{\circ}$