Đáp án:
a)
Xét ΔABD và ΔECD có:
+ góc BAD = góc CED = 90 độ
+ góc ADB = góc EDC (đối đỉnh)
=> ΔABD ~ ΔECD (g-g)
b)
$\begin{array}{l}
Do:\Delta ABD \sim \Delta ECD\\
\Rightarrow \dfrac{{DA}}{{DE}} = \dfrac{{DB}}{{DC}}\\
\Rightarrow DA.DC = DB.DE
\end{array}$
c)
$\begin{array}{l}
Theo\,Pytago:B{C^2} = A{B^2} + A{C^2}\\
\Rightarrow B{C^2} = {3^2} + {4^2} = 25\\
\Rightarrow BC = 5\left( {cm} \right)\\
Theo\,t/c:\dfrac{{AD}}{{AB}} = \dfrac{{DC}}{{BC}}\\
\Rightarrow \dfrac{{AD}}{3} = \dfrac{{DC}}{5} = \dfrac{{AD + DC}}{{3 + 5}} = \dfrac{4}{8} = \dfrac{1}{2}\\
\Rightarrow \left\{ \begin{array}{l}
AD = 1,5\left( {cm} \right)\\
DC = 2,5\left( {cm} \right)
\end{array} \right.\\
\Rightarrow \dfrac{{{S_{ABD}}}}{{{S_{BCD}}}} = \dfrac{{\dfrac{1}{2}.AB.AD}}{{\dfrac{1}{2}.AB.CD}} = \dfrac{{AD}}{{CD}} = \dfrac{{1,5}}{{2,5}} = \dfrac{3}{5}\\
d)Xet\,\Delta DBC;\Delta DAE\\
+ \dfrac{{DB}}{{DA}} = \dfrac{{DC}}{{DE}}\\
+ \widehat {BDC} = \widehat {EDA}\\
\Rightarrow \Delta DBC \sim \Delta DAE\left( {g - g} \right)\\
\Rightarrow \widehat {DBC} = \widehat {DAE}\\
Do:\Delta ABD \sim \Delta ECD\\
\Rightarrow \widehat {DBA} = \widehat {DCE}\\
\widehat {DBC} = \widehat {DBA}\\
\Rightarrow \widehat {CAE} = \widehat {ACE}
\end{array}$
=> tam giác ACE cân tại E
