Mình cũng k chắc trình bày như này có đúng k nữa :'< Tại lâu k làm toán lớp 6 :'<
$\text{a) Có: OC là tia p/g của $\widehat{AOB}$ (gt)}$
$\text{⇒ $\widehat{AOC}$ = $\widehat{BOC}$ = $\dfrac{\widehat{AOB}}{2}$ = $\dfrac{140^o}{2}$ = $70^{o}$}$
$\text{Trên nửa mặt phẳng bờ chứa tia OA có:}$
$\text{$\widehat{AOC}$ = $70^{o}$; $\widehat{AOD}$ = $180^{o}$}$
$\text{⇒ $\widehat{AOC}$ < $\widehat{AOD}$}$
$\text{⇒ Tia OC nằm giữa OA và OD}$
$\text{Có: $\widehat{AOC}$ + $\widehat{DOC}$ = $180^{o}$ (kề bù)}$
$\text{⇒ $\widehat{DOC}$ = $180^{o}$ - $\widehat{AOC}$}$
$\text{⇒ $\widehat{DOC}$ = $180^{o}$ - $70^{o}$}$
$\text{⇒ $\widehat{DOC}$ = $110^{o}$}$
$\text{b) Có: $\widehat{AOE}$ = $\dfrac{5}{7}$$\widehat{AOB}$}$
$\text{⇒ $\widehat{AOE}$ = $\dfrac{5}{7}$ . $140^{o}$ = $100^{o}$}$
$\text{Trên nửa mặt phẳng bờ chứa tia OA có:}$
$\text{$\widehat{AOE}$ = $100^{o}$; $\widehat{AOD}$ = $180^{o}$}$
$\text{⇒ $\widehat{AOE}$ < $\widehat{AOD}$}$
$\text{⇒ Tia OE nằm giữa OA và OD}$
$\text{Có: $\widehat{AOE}$ + $\widehat{EOD}$ = $180^{o}$ (kề bù)}$
$\text{⇒ $\widehat{EOD}$ = $180^{o}$ - $\widehat{AOE}$}$
$\text{⇒ $\widehat{EOD}$ = $180^{o}$ - $100^{o}$}$
$\text{⇒ $\widehat{EOD}$ = $80^{o}$}$
$\text{Có: $\widehat{AOB}$ + $\widehat{BOD}$ = $180^{o}$ (kề bù)}$
$\text{⇒ $\widehat{BOD}$ = $180^{o}$ - $\widehat{AOB}$}$
$\text{⇒ $\widehat{BOD}$ = $180^{o}$ - $140^{o}$}$
$\text{⇒ $\widehat{BOD}$ = $40^{o}$}$
$\text{Có: $\widehat{BOD}$ + $\widehat{EOD}$ = $\widehat{EOD}$}$
$\text{⇒ $\widehat{EOD}$ = $\widehat{EOD}$ - $\widehat{BOD}$}$
$\text{⇒ $\widehat{EOD}$ = $80^{o}$ - $40^{o}$}$
$\text{⇒ $\widehat{EOD}$ = $40^{o}$}$
$\text{Có: $\widehat{BOD}$ = $\widehat{EOB}$ = $40^{o}$}$
$\text{⇒ OB là p/g của $\widehat{EOD}$}$
