Xét 2 tam giác ADB và BCD có:
DAB = DBC
BD chung
ABD = BDC (AB//DC,So le trong)
=> \(\Delta ABD\) ~ \(\Delta BDC\) (g.c.g)
=> \(\dfrac{AB}{DB}=\dfrac{DB}{DC}=>DB^2=AB.DC=>DB=\sqrt{324}=>DB=18cm\)
AB // CD(gt)\(\Rightarrow\)\(\widehat{ABD}\)=\(\widehat{BDC}\) (2 góc so le trong)
Xét \(\Delta\)ABD và \(\Delta\)BDC
có : \(\widehat{ABD}\)=\(\widehat{BDC}\)( CMT)
\(\widehat{DAB}\)=\(\widehat{DBC}\) (gt)
Do đó :\(\Delta\)ABD ~ \(\Delta\)BDC(gg)
\(\Rightarrow\)\(\dfrac{AB}{BD}=\dfrac{BD}{DC}\)(định nghĩa 2 tam giác đồng dạng)
\(\Rightarrow\)BD2 = AB. DC
\(\Rightarrow\)BD2 = 12.27
\(\Rightarrow\)BD2 = 324
\(\Rightarrow\)BD2 = 182
\(\Rightarrow\) BD = 18 (cm)
