Kẻ $ab$ qua $C$ sao cho $ab//AB$
$→\widehat{ABC}+\widehat{BCa}=180^\circ$ (trong cùng phía)
mà $\widehat{ABC}=120^\circ$
$→\widehat{BCa}=180^\circ-120^\circ=60^\circ$
Lại có: $\widehat{BCa}+\widehat{BCD}+\widehat{DCb}=180^\circ$
$→\widehat{DCb}=180^\circ-\widehat{BCa}-\widehat{BCD}=180^\circ-60^\circ-75^\circ=45^\circ$
$→\widehat{DCb}+\widehat{EDC}=135^\circ+45^\circ=180^\circ$
mà 2 góc ở vị trí trong cùng phía
$→DE//ab$ mà $AB//ab$
$→AB//DE$
