Cho biết $ABCD$ là hình chữ nhật. Tìm $x.$                    

$\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}}$

$AD.AE = AB.AC$

$\dfrac{{AD}}{{DB}} = \dfrac{{DE}}{{BC}}$

$DE.AD = AB.BC$

$\Delta ABC = \Delta A'B'C' \Rightarrow \Delta ABC \backsim \Delta A'B'C'$

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

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

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

$\dfrac{{LC}}{{LB}} = \dfrac{{LK}}{{LA}}$  

$\dfrac{{IB}}{{IK}} = \dfrac{{IA}}{{ID}}$

$\dfrac{{IB}}{{ID}} = \dfrac{{IA}}{{IK}}$      

$\dfrac{{KA}}{{KL}} = \dfrac{{KD}}{{KC}}$

$k$

$\dfrac{1}{k}$         

${k^2}$

$2k$

$\dfrac{{HG}}{{MN}} = \dfrac{7}{2}$   

$\dfrac{{{S_{\Delta MNP}}}}{{{S_{\Delta HGK}}}} = \dfrac{2}{7}$

$\dfrac{{{S_{\Delta MNP}}}}{{{S_{\Delta HGK}}}} = \dfrac{{49}}{4}$     

$\dfrac{{NP}}{{GK}} = \dfrac{5}{7}$

$3\dfrac{1}{4}$

$6$    

$6\dfrac{1}{4}$       

$6\dfrac{2}{3}$

$\dfrac{1}{3}$

$\dfrac{1}{4}$

$\dfrac{1}{8}$

$1$