a, Xét ΔABD và ΔBDC có :
\(\widehat{A}=\widehat{DBC}\left(gt\right)\)
\(\widehat{ABD}=\widehat{BDC}\left(AB//CD;slt\right)\)
\(\Rightarrow\Delta ABD\sim\Delta BDC\left(g-g\right)\)
b, Ta có : \(\Delta ABD\sim\Delta BDC\left(cmt\right)\)
\(\Rightarrow\dfrac{AB}{BD}=\dfrac{BD}{DC}\)
hay \(BD^2=AB.DC=12.28,5=342\)
\(\Rightarrow BD=\sqrt{342}\left(cm\right)\)