Giải thích các bước giải:

a.Ta có:

$S_{ABC}=\dfrac12AD\cdot AB$

$S_{ADC}=\dfrac13AD\cdot CD$

Vì $AB=\dfrac13CD$

$\to S_{ABC}=\dfrac13S_{ADC}$

b.Ta có:

$S_{ABC}=\dfrac13S_{ADC}=\dfrac13S_{DBC}$

$\to$Khoảng cách từ $A$ đến $CB=\dfrac13$ khoảng cách từ $D$ đến $BC$

$\to S_{EAC}=\dfrac13S_{DEC}$

$\to S_{DEC}-S_{EAC}=\dfrac23S_{DEC}$

$\to S_{ADC}=\dfrac23S_{DEC}$

$\to 3S_{ABD}=\dfrac23S_{DEC}$

$\to S_{ADB}=\dfrac29S_{DEC}$

Ta có: $S_{ADC}=3S_{BCD}=3S_{ADB}$

$\to S_{ABCD}=S_{ABD}+S_{BCD}=4S_{ABD}$

$\to S_{DEC}=S_{ABE}+S_{ABCD}=7+4S_{ABD}$

$\to S_{ADB}=\dfrac29(7+4S_{ABD})$

$\to S_{ADB}=\dfrac{14}9+\dfrac89S_{ABD}$

$\to \dfrac19S_{ABD}=\dfrac{14}9$

$\to S_{ABD}=14$

$\to S_{ABCD}=4\cdot 14=56$