[tex]\it \left.\begin{aligned}\widehat{BAC}=\widehat{ADC}=90^o\\ \\ \hat C\ -\ unghi\ comun \end{aligned}\right\} \stackrel{UU}{\Longrightarrow}\ \Delta ABC\sim \Delta DAC\\ \\ \\ \left.\begin{aligned}\widehat{DAB}=\widehat{C}\ (au\ acela\c{s}i\ complement, \hat B)\\ \\ \widehat{BDA}\ =\widehat{CED}=90^o \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rule{0.2}{0.1} \end{aligned}\right\} \stackrel{UU}{\Longrightarrow}\ \Delta ABD\sim \Delta CDE[/tex]
c)
[tex]\it \Delta ABC\sim \Delta DAC \Rightarrow \dfrac{BC}{AC}=\dfrac{AC}{CD}\Rightarrow \dfrac{BC}{16}=\dfrac{16}{12,8} \Rightarrow BC=\dfrac{16\cdot16}{12,8} \Rightarrow \\ \\ \Rightarrow BC=\dfrac{\ \ 256^{(10}}{12,8}=\dfrac{2560}{128}=20\ cm[/tex]
[tex]\it \begin{cases}BA\perp AC\\ \\ DE\perp AC \end{cases} \Rightarrow DE||BA\ \stackrel{T.Thales}{\Longrightarrow}\ \dfrac{EC}{AC}=\dfrac{CD}{BC} \Rightarrow EC=\dfrac{AC\cdot CD}{BC}=\\ \\ \\ =\dfrac{12,8\cdot16^{(2}}{20}=\dfrac{12,8\cdot8}{10}=\dfrac{102,4}{10}=10,24\ cm[/tex]
