Ta có: \(\widehat{ABC}=\widehat{BCx}=40^0\) (2 góc sole trong)
\(\widehat{EDC}=\widehat{DCx}=65^0\) (2 góc sole trong)
mà \(\widehat{BCx}+\widehat{DCx}=\widehat{BCD}\)
\(\Rightarrow\) \(\widehat{BCD}=65^0+40^0=105^0\)
+)\(\widehat{E_1}=\widehat{A_1}=90^o\)
=>EA⊥AB
EA⊥ED
=>AB//ED
+)Kẻ Cx//AB
=>Cx//ED(vì cùng song song AB)
=>\(\widehat{DCx}=\widehat{FDC}=65^o\) (so le trong)
+)Cx//AB
\(=>\widehat{BCx}=\widehat{ABC}=40^o\)(so le trong)
+)\(\widehat{BCD}=\widehat{DCx}+\widehat{BCx}=65^o+40^o=105^o\)
vì Cx//AB
=> góc ABC=BCx=40 độ (2 góc sole trong)
vì Cx//DE
=> góc FDC=DCx=65 độ (2 góc sole trong)
mà góc BCD = BCx + DCx
góc BCD = 40 + 65
góc BCD = 105 độ
