Question

Cho hình vẽ ; biết \(\widehat{AMN}\)\(=\)\(2\widehat{BMN}\)

\(\widehat{MND}-\widehat{MNC}=60^0\)

CMR: AB // CD

Answer 1

Giải:

Theo đề ra, ta có:

\(\widehat{AMN}=2\widehat{BMN}\)

Mà \(\widehat{AMN}+\widehat{BMN}=180^0\) (Hai góc kề bù)

\(\Leftrightarrow3\widehat{BMN}=180^0\)

\(\Leftrightarrow\widehat{BMN}=\dfrac{180^0}{3}=60^0\)

\(\Leftrightarrow\widehat{AMN}=180^0-60^0=120^0\)

Lại có:

\(\widehat{MND}-\widehat{MNC}=60^0\)

Mà \(\widehat{MND}+\widehat{MNC}=180^0\) (Hai góc kề bù)

\(\Leftrightarrow\widehat{MND}=\dfrac{180^0+60^0}{2}=120^0\)

\(\Leftrightarrow\widehat{MNC}=180^0-120^0=60^0\)

Vì \(\widehat{BMN}=\widehat{MNC}\left(=60^0\right)\)

\(\Leftrightarrow\) \(AB//CD\) (Vì có hai góc so le trong bằng nhau)

\(\Rightarrowđpcm\)

Chúc bạn học tốt!