1,
Ta có :
AD là tia phân giác \(\widehat{BAC}\)
=> \(\dfrac{AB}{AC}=\dfrac{BD}{CD}\)
=> \(\dfrac{5}{8,5}=\dfrac{3}{CD}\)
=> \(CD=\dfrac{3.8,5}{5}=5,1\left(cm\right)\)
a) △ABD và △BDC có: \(\dfrac{AB}{BD}=\dfrac{BD}{DC}=\dfrac{AD}{BC}\left(=\dfrac{1}{2}\right)\)
\(\Rightarrow\)△ABD∼△BDC (c-c-c).
b) △ABD∼△BDC \(\Rightarrow\widehat{ABD}=\widehat{BDC}\)
\(\Rightarrow\)AB//CD \(\Rightarrow\)ABCD là hình thang.