Question

giúp mik vs ạ, đầy đủ cả hình 

Answer 1

Ta có: \(\widehat{A}=180^0-\left(\widehat{B}+\widehat{C}\right)=180^0-\left(80^0+40^0\right)=60^0\)

Do \(BD\) là phân giác nên \(\widehat{ABD}=\dfrac{1}{2}\widehat{B}=\dfrac{1}{2}.80^0=40^0\)

\(\Rightarrow\widehat{ADB}=180^0-\left(\widehat{A}+\widehat{ABD}\right)=180^0-\left(60^0+40^0\right)=80^0\)