Cho \(\Delta ABC\), đường phân giác góc $B$ cắt $AC$ tại $D$ và cho biết $AB = 10{\rm{ }}cm,{\rm{ }}BC = 15{\rm{ }}cm,{\rm{ }}AD = 6{\rm{ }}cm.$ Tính $AC = $ ?

\(\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}}\)

\(AD.AE = AB.AC\)

\(\dfrac{{AD}}{{DB}} = \dfrac{{DE}}{{BC}}\)

\(DE.AD = AB.BC\)

\(\Delta ABC = \Delta A'B'C' \Rightarrow \Delta ABC \backsim \Delta A'B'C'\)

\(\widehat A = \widehat {A'},\;\widehat B = \widehat {B'} \Rightarrow \Delta ABC \backsim \Delta A'B'C'\)

\(\dfrac{{AB}}{{A'B'}} = \dfrac{{BC}}{{B'C'}} \Rightarrow \Delta ABC \backsim \Delta A'B'C'\)       

\(\Delta ABC = \Delta A'B'C' \Rightarrow {S_{\Delta ABC}} = {S_{\Delta A'B'C'}}\)

\(\dfrac{{LC}}{{LB}} = \dfrac{{LK}}{{LA}}\)  

\(\dfrac{{IB}}{{IK}} = \dfrac{{IA}}{{ID}}\)

\(\dfrac{{IB}}{{ID}} = \dfrac{{IA}}{{IK}}\)      

\(\dfrac{{KA}}{{KL}} = \dfrac{{KD}}{{KC}}\)

\(k\)

\(\dfrac{1}{k}\)         

\({k^2}\)

\(2k\)

\(\dfrac{{HG}}{{MN}} = \dfrac{7}{2}\)   

\(\dfrac{{{S_{\Delta MNP}}}}{{{S_{\Delta HGK}}}} = \dfrac{2}{7}\)

\(\dfrac{{{S_{\Delta MNP}}}}{{{S_{\Delta HGK}}}} = \dfrac{{49}}{4}\)     

\(\dfrac{{NP}}{{GK}} = \dfrac{5}{7}\)

\(3\dfrac{1}{4}\)

\(6\)    

\(6\dfrac{1}{4}\)       

\(6\dfrac{2}{3}\)

\(\dfrac{1}{3}\)

\(\dfrac{1}{4}\)

\(\dfrac{1}{8}\)

\(1\)