Đáp án:
$\begin{array}{l}
1)\widehat {ABE} = \widehat {EBC} = \widehat {ACD} = \widehat {DCB} = \dfrac{1}{2}\widehat {ABC}\\
\Rightarrow DE//BC\\
\Rightarrow \dfrac{{AD}}{{AB}} = \dfrac{{DE}}{{BC}}\left( {Theo\,Talet} \right)\\
\Rightarrow \dfrac{{AD}}{{AB}} = \dfrac{{10}}{{16}} = \dfrac{5}{8}\\
\Rightarrow \dfrac{{AD}}{{BD}} = \dfrac{5}{3}\\
Theo\,t/c:\dfrac{{AD}}{{BD}} = \dfrac{{AC}}{{BC}}\\
\Rightarrow \dfrac{5}{3} = \dfrac{{AB}}{{16}}\\
\Rightarrow AB = \dfrac{{5.16}}{3} = \dfrac{{80}}{3}\left( {cm} \right)\\
Vậy\,AB = \dfrac{{80}}{3}cm\\
2)Theo\,Talet:\\
\dfrac{{CD}}{{CB}} = \dfrac{{CE}}{{AC}} \Rightarrow \dfrac{5}{{5 + 7,5}} = \dfrac{{CE}}{{10}}\\
\Rightarrow CE = 4\left( {cm} \right)\\
\Rightarrow AE = AC - CE = 10 - 4 = 6\left( {cm} \right)\\
Theo\,t/c:\dfrac{{BD}}{{AB}} = \dfrac{{CD}}{{AC}}\\
\Rightarrow \dfrac{{7,5}}{{AB}} = \dfrac{5}{{10}}\\
\Rightarrow AB = 15\left( {cm} \right)\\
\dfrac{{DE}}{{AB}} = \dfrac{{CD}}{{BC}}\\
\Rightarrow DE = \dfrac{5}{{12,5}}.15 = 6\left( {cm} \right)\\
Vậy\,AE = 6cm;EC = 4cm;DE = 6cm
\end{array}$
