Do BC//AD (gt) => \(\widehat{BCA}=\widehat{CAD}\left(slt\right)\)
Xét ΔABC và ΔACD có: \(\widehat{BCA}=\widehat{CAD}\left(cmt\right)\)
\(\widehat{ABC}=\widehat{ACD}\left(gt\right)\)
=> ΔABC ~ ΔDCA (g-g)
=> \(\frac{AC}{DA}=\frac{BC}{AC}\)
=> AC2 = AD.BC = 12.27 = 324 => AC = 18 (m)