a, Ta có: \(\widehat{mAx}=\widehat{mBy}=120^0\left(GT\right)\)
Mà 2 góc đang ở vị trí so le trong nên:
=> Ax // By
b, Ta có: \(\widehat{mBC}+\widehat{CBy}+\widehat{mBy}=360^0\)
\(\Rightarrow\widehat{CBy}=360^0-\widehat{mBC}-\widehat{mBy}\)
\(\Rightarrow\widehat{CBy}=360^0-90^0-120^0\)
\(\Rightarrow\widehat{CBy}=150^0\)
Lại có : \(\widehat{CBy}=\widehat{BCz}=150^0\)
Mà 2 góc đag ở vị trí so le trong nên:
\(\Rightarrow By//Cz\)
c, Ta có: \(\widehat{mAx}+\widehat{xAB}=180^{^0}\) ( kề bù)
=> \(\widehat{xAB}=180^0-\widehat{mAx}=180^0-120^0=60^0\)
Mà \(\widehat{A1}=\widehat{A2}=\widehat{\frac{BAx}{2}}=\frac{60^0}{2}=30^0\)
Ta lại có: \(\widehat{mAe}+\widehat{eAt}+\widehat{tAx}+\widehat{xAm}=360^0\)
\(\Rightarrow\widehat{eAt}=360^0-\widehat{mAe}-\widehat{mAx}-\widehat{xAt}=360^0-120^0-120^0-30^0=90^0\)
\(\Rightarrow At\perp Ae\left(đpcm\right)\)
