$\triangle$ ABC có $\widehat{BAC}$ = $100^{o}$ ; $\widehat{ABC}$ = $40^{o}$
$\Rightarrow$ $\widehat{ACB}$ = $40^{o}$
Ta có $\widehat{xAC}$ = $180^{o}$ - $ \widehat{BAC}$ (2 góc kề bù)
= $180^{o}$ - $100^{o}$ = $80^{o}$
$\Rightarrow$ $\widehat{yAC}$ = $\frac{\widehat{xAC}}{2}$ = $\frac{80^o}{2}$ = $40^{o}$
Ta có $\widehat{yAC}$ = $\widehat{ACB}$ = $40^{o}$ mà chúng ở vị trí so le trong
$\Rightarrow$ Ay║ BC