Đáp án:
a) Theo Pytago:
$\begin{array}{l}
A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow A{C^2} = {5^2} - {3^2} = 16\\
\Rightarrow AC = 4\left( {cm} \right)\\
\sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{4}{5}\\
\Rightarrow \widehat B = {53^0}\\
\Rightarrow \widehat C = {90^0} - {53^0} = {37^0}\\
Vậy\,AC = 4cm;\widehat B = {53^0};\widehat C = {37^0}\\
b)Theo\,t/c:\\
\dfrac{{AD}}{{AB}} = \dfrac{{DC}}{{BC}}\\
\Rightarrow \dfrac{{AD}}{3} = \dfrac{{DC}}{5} = \dfrac{{AD + DC}}{{3 + 5}} = \dfrac{4}{8} = \dfrac{1}{2}\\
\Rightarrow \left\{ \begin{array}{l}
AD = 1,5\left( {cm} \right)\\
DC = 2,5\left( {cm} \right)
\end{array} \right.
\end{array}$
c)
Kẻ FM ⊥AB tại M
Xét ΔFME và ΔDAE có:
+ góc FME = góc DAE = 90 độ
+ góc FEM = góc DEA (đối đỉnh)
=> ΔFME ~ ΔDAE (g-g)
$\begin{array}{l}
\Rightarrow \dfrac{{FM}}{{AD}} = \dfrac{{ME}}{{AD}} = \dfrac{1}{2}\\
\Rightarrow FM = \dfrac{1}{2}.\dfrac{3}{2} = \dfrac{3}{4}\left( {cm} \right)\\
\Rightarrow {S_{BEF}} = \dfrac{1}{2}.FM.BE = \dfrac{1}{2}.\dfrac{3}{4}.\dfrac{1}{4}.3 = \dfrac{9}{{32}}\\
{S_{BEDC}} = {S_{ABC}} - {S_{ADE}}\\
= \dfrac{1}{2}.3.4 - \dfrac{1}{2}.\dfrac{3}{4}.3.\dfrac{3}{2}\\
= 6 - \dfrac{{27}}{{16}}\\
= \dfrac{{69}}{{16}}\\
\Rightarrow \dfrac{{{S_{BEF}}}}{{{S_{BEDC}}}} = \dfrac{9}{{32}}:\dfrac{{69}}{{16}} = \dfrac{3}{{46}}
\end{array}$
