AECF là hình bình hành (Vì có AE,FC song song và bằng nhau), suy ra: // .

Khi đó, ta có: $\widehat{AFD}=\widehat{ECF}$  (hai góc đồng vị)

                         $\widehat{CEB}=\widehat{ECF}$(hai góc so le trong)

Từ đó:$\widehat{AFD}=\widehat{CEB}$

Xét $\Delta ADF$  và $\Delta CBE$ , ta có:

                                            $\widehat{B}=\widehat{D}={90}^0$              

                                                $\widehat{AFD}=\widehat{CEB}$        

Do vậy:$\Delta ADF~\Delta CBE$ (g.g)

Xét $\Delta ABC$  và $\Delta HBA$ , ta có:

                                                     $\widehat{A}=\widehat{H}={90}^0$     

                                                        $\widehat{B}$ chung

Do đó: $\Delta ABC ~\Delta HBA$  (g.g)

 $\Rightarrow \frac{AB}{HB}=\frac{BC}{BA}$

$\Rightarrow AB.BA=HB.BC$ hay $A{B}^2=BC.BH$

Áp dụng định lý Py-ta-go vào tam giác vuông ABC:

                                                           $A{B}^2+A{C}^2=B{C}^2$                  

                                                           $\Rightarrow {15}^2+{20}^2=B{C}^2$              

                                                           $\Rightarrow BC=25cm$                 .

Theo a, ta có: $\frac{AB}{HB}=\frac{BC}{BA}$  hay $\frac{15}{HB}=\frac{25}{15}$

 $\Rightarrow HB=\frac{15.15}{25}=9cm.$

$CH=BC-HB=25-9=16cm.$

Vậy $HB=9cm,CH=16cm.$

Xét $\Delta ADB$  và $\Delta BCD$  có:

                                                         $\widehat{ABD}=\widehat{BDC}$  (hai góc so le trong)

                                                        $\widehat{DBC}=\widehat{DAB}$

Do đó: $\Delta ADB ~\Delta BCD$  (g.g)

Vì $\Delta ADB ~\Delta BCD$  nên  $\frac{AD}{BC}=\frac{AB}{BD}=\frac{DB}{CD}$

Hay $\frac{2,5}{BC}=\frac{3}{6}=\frac{6}{CD}$

 $\Rightarrow BC=5cm,CD=12cm.$