Cho hai góc kề bù 

- Hướng dẫn giải

Ta có: $\widehat{aOb}-\widehat{bOc}={120}^0\Rightarrow \widehat{aOb}={120}^0+\widehat{bOc}$

Vì $\widehat{aOb}$ và $\widehat{bOc}$ là hai góc kề bù nên  $\widehat{aOb}+\widehat{bOc}={180}^0$

 $\Rightarrow {120}^0+\widehat{bOc}+\widehat{bOc}={180}^0\Rightarrow 2\widehat{bOc}={60}^0\Rightarrow \widehat{bOc}={30}^0$

$\Rightarrow \widehat{aOb}={150}^0$ 

Vì Od nằm trong góc $\widehat{aOb}$ nên $\widehat{aOd}+\widehat{dOb}=\widehat{aOb}$

$\Rightarrow {60}^0+\widehat{dOb}={150}^0\Rightarrow \widehat{dOb}={90}^0$

  Vậy $Ob\perp Od$ (đpcm)