Bài 1:
a) $\widehat{N}=\widehat{O}$
mà $\widehat{N}=90^o$
$→\widehat{O}=90^o$
$→AO⊥BO$
mà $MQ⊥BO$
$→AO//MQ$ mà $MN⊥AO$
$→MQ⊥MN$
$→\widehat{NMQ}=90^o$
b) $MN⊥AO$ mà $BO⊥AO$
$→MN//BO$
$→\widehat{AMN}=\widehat{B}=50^o$ (đồng vị)
$→\widehat{QMB}=180^o-90^o-50^o=40^o$
Bài 2:
Kẻ $EH$ về phía bên phải sao cho $EH//AB$
$→EH//AB//CD$
$→\widehat{A}+\widehat{AEH}=180^o$ (trong cùng phía)
mà $\widehat{A}=140^o$
$→\widehat{AEH}=180^o-140^o=40^o$
$→\widehat{CEH}=120^o-40^o=80^o$
$EH//CD$
$→\widehat{CEH}+\widehat{ECD}=180^o$ (trong cùng phía)
mà $\widehat{CEH}=80^o$
$→\widehat{ECD}=100^o$
b) $AB//EH$
$→\widehat{EAB}+\widehat{AEH}=180^o$ (trong cùng phía)
$EH//CD$
$→\widehat{ECD}+\widehat{CEH}=180^o$ (trong cùng phía)
Từ hai điều trên
$→\widehat{EAB}+\widehat{AEH}+\widehat{CEH}+\widehat{ECD}=180^o+180^o=360^o$
$→\widehat{EAB}+\widehat{AEC}+\widehat{ECD}=360^o$ (ĐPCM)
