Bài 3:
$AB∩CD≡{E}$
$⇒\widehat{AED}=\widehat{BEC}=64^o$ (đối đỉnh)
Ta có:
$\widehat{AED}+\widehat{BED}=180^o$
mà $\widehat{AED}=64^o$
$⇒\widehat{BED}=180^o-64^o=116^o$
$⇒\widehat{BED}=\widehat{AEC}=116^o$ (đối đỉnh)
Bài 4:
Ta có:
$\widehat{BOC}+\widehat{AOC}+\widehat{AOD}+\widehat{BOD}=360^o$
$⇒\widehat{AOD}+240^o=360^o$
$⇒\widehat{AOD}=360^o-240^o=120^o$
$⇒\widehat{AOC}=180^o-120^o=60^o$
mà $\widehat{BOC}=\widehat{AOD}$ (đối đỉnh)
$\widehat{AOC}=\widehat{BOD}$ (đối đỉnh)
$⇒\widehat{BOC}=\widehat{AOD}=120^o$
$\widehat{AOC}=\widehat{BOD}=60^o$
