a,
$\Delta AEC$ và $\Delta ADB$ có:
$\widehat{AEC}=\widehat{ADB}=90^o$
$\widehat{A}$ chung
$\to\Delta AEC\backsim\Delta ADB$ (g.g)
$\to \dfrac{AE}{AC}=\dfrac{AD}{AB}$
$\to AE.AB=AD.AC$
b,
$\dfrac{AE}{AC}=\dfrac{AD}{AB}\to \dfrac{AE}{AD}=\dfrac{AC}{AB}$
Mà $\widehat{A}$ chung nên $\Delta ADE\backsim \Delta ABC$ (c.g.c)
$\to \widehat{ADE}=\widehat{ABC}$
c,
$\widehat{ACE}=90^o-\widehat{A}=90^o-60^o=30^o$
$\Delta AEC$ vuông tại $E$, $AE$ đối diện góc $30^o$ nên $AE=\dfrac{1}{2}AC$
$\to \dfrac{AE}{AC}=\dfrac{1}{2}$
$\to \Delta ADE\backsim\Delta ABC$ với $k=\dfrac{1}{2}$
$\to$ $\dfrac{S_{ADE}}{S_{ABC}}=k^2=\dfrac{1}{4}$
Vậy $S_{ADE}=\dfrac{S_{ABC}}{4}=37(cm^2)$