Do $\widehat{EDC}$ = $\widehat{ACB}$
Và 2 góc so le trong
$\Rightarrow$ DE // BC
Xét $\triangle$ABC, có:
$\widehat{BAC}$ + $\widehat{B}$ + $\widehat{C}$ = $180^o$
$\Rightarrow$ $\widehat{C}$= $180^o$ - $80^o$ - $50^o$ = $50^o$
$\Rightarrow$ $\widehat{EDC}$ = $50^o$ (1)
Ta có: $\widehat{DAC}$=$180^o$
$\Rightarrow$ $\widehat{DAB}$ + $\widehat{BAC}$ = $180^o$
$\Rightarrow$ $\widehat{DAB}$ = $180^o$ -$\widehat{BAC}$ =$180^o$-$80^o$=$100^o$
$\Rightarrow$ 2.$\widehat{MAB}$ = $100^o$ ( AM là phân giác của $\widehat{BAD}$)
$\Rightarrow$ $\widehat{MAB}$ = $50^o$
Từ (1) và (2)$\Rightarrow$ $\widehat{MAB}$ = $\widehat{MAB}$ = $50^o$
$\Rightarrow$ AM // DE ( 2 góc so le trong bằng nhau)
