Question

[TRí Minh Thái]

mn lm giup em bai 7,8 vs a

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 7: Kẻ CF//AB//DE(CF,AB cùng nằm trên nửa mặt phẳng bờ AC)

CF//AB

=>\(\widehat{BAC}+\widehat{ACF}=180^0\)

CF//DE

=>\(\widehat{FCD}+\widehat{CDE}=180^0\)

\(\widehat{BAC}+\widehat{ACF}+\widehat{FCD}+\widehat{CDE}\)

\(=180^0+180^0=360^0\)

=>\(\widehat{BAC}+\widehat{ACD}+\widehat{CDE}=360^0\)

Bài 8:

Qua C, kẻ FH//AB//DE

FC//AB

=>\(\widehat{ABC}+\widehat{FCB}=180^0\)

=>\(\widehat{FCB}=68^0\)

Ta có: CH//DE

=>\(\widehat{CDE}+\widehat{DCH}=180^0\)

=>\(\widehat{DCH}=180^0-133^0=47^0\)

Ta có: \(\widehat{BCF}+\widehat{BCD}+\widehat{HCD}=180^0\)

=>\(\widehat{BCD}+47^0+68^0=180^0\)

=>\(\widehat{BCD}=65^0\)