Xét ∆ABD và ∆BDC có:
+) \(\widehat{DAB}\) = \(\widehat{DBC}\) (gt)
+) \(\widehat{ABD}\) = \(\widehat{BDC}\) (Hai góc so le trong)
\(=> ∆ABD ∽ ∆BDC\) (g-g)
=> \(\dfrac{AB}{BD}\) = \(\dfrac{BD}{DC}\) (tính chất hai tam giác đồng dạng)
=> BD2 = AB.DC
\( =>BD = \sqrt {AB.DC} = \sqrt {12,5.28,5} \) \( \approx 18,87 cm\)
Xét \(\Delta ABC\) và \(\Delta BDC\), ta có:
\(\widehat{DAB}=\widehat{DBC}\left(gt\right)\)
\(\widehat{ABD}=\widehat{BDC}\left(AB//CD\right)\)
\(\Rightarrow\Delta ABC\sim\Delta BDC\left(g.g\right)\)
\(\Rightarrow\frac{AB}{BD}=\frac{BD}{DC}\Rightarrow BD^2=AB.DC\)
\(\Rightarrow BD=\sqrt{AB.DC}=\sqrt{12,5.28,5}\)
\(\Rightarrow BD\approx18,87cm\) hay \(x\approx18,87cm\)