a: Các cặp góc đồng vị bằng nhau là \(\widehat{nAm};\widehat{AED}\); \(\widehat{nAB};\widehat{AEt}\); \(\widehat{pBA};\widehat{pDx}\); \(\widehat{pBQ};\widehat{CDE}\); \(\widehat{CAB};\widehat{zEt}\); \(\widehat{ABC};\widehat{BDx}\); \(\widehat{qBD};\widehat{yDE}\); \(\widehat{mAC};\widehat{zED}\)
b: Ta có: \(\widehat{BAC}=\widehat{zEt}\)
=>\(\widehat{BAC}=45^0\)
Xét ΔCAB có \(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^0\)
=>\(\widehat{ACB}=180^0-45^0-37^0=180^0-82^0=98^0\)
Ta có: AB//CD
=>\(\widehat{CDE}=\widehat{CBA}\)(hai góc so le trong)
=>\(\widehat{CDE}=37^0\)
c: Qua C, kẻ tia CM nằm giữa hai tia CB và CE sao cho CM//AB//DE
CM//DE
=>\(\widehat{MCE}=\widehat{CED}\)(hai góc so le trong)
=>\(\widehat{MCE}=45^0\)
Ta có: CM//AB
=>\(\widehat{ABC}=\widehat{MCB}\)(hai góc so le trong)
=>\(\widehat{MCB}=37^0\)
\(\widehat{BCE}=\widehat{MCB}+\widehat{MCE}=37^0+45^0=82^0\)
=>bạn Nam nói đúng
