[tex]\it [AD\ -\ bisectoare\ \stackrel{T.bisec.}{\Longrightarrow}\ \dfrac{CD}{DB}=\dfrac{AC}{AB} \Rightarrow \dfrac{CD}{DB}=\dfrac{\ 12^{(4}}{8} \Rightarrow \dfrac{CD}{DB}=\dfrac{3}{2}=1,5\ \ \ (1)\\ \\ \\ DE||AB\ \stackrel{T.Thales}{\Longrightarrow}\ \dfrac{CE}{EA}=\dfrac{CD}{DB}\ \stackrel{(1)}{\Longrightarrow}\ \dfrac{CE}{EA}=1,5 \Rightarrow CE=1,5 EA\ \ \ \ \ (2)\\ \\ \\ CE+EA=AC\ \stackrel{(2)}{\Longrightarrow}1,5EA+EA+12 \Rightarrow 2,5EA=12\bigg|_{\cdot4}\Rightarrow[/tex]
[tex]\it \Rightarrow 10EA=48\bigg|_{:10} \Rightarrow EA=4,8cm \Rightarrow AE=4,8cm\\ \\ \\ CE=AC=AE=12-4,8=7,2cm[/tex]
[tex]\it \left.\begin{aligned}DE||AB \Rightarrow \widehat{DAB}=\widehat{ADE}\ (alterne\ interne)\\ \\ \\ Dar,\ [AD-bisectoare \Rightarrow \widehat{DAB}=\widehat{EAD} \ \ \ \end{aligned} \right\} \Rightarrow\widehat{EAD}=\widehat{ADE} \Rightarrow\\ \\ \\ \Rightarrow \Delta EAD-isoscel,\ DE=AE=4,8cm[/tex]
