Giải thích các bước giải:
a.Ta có : $EF//BC\to\dfrac{AF}{AC}=\dfrac{AE}{AB}$
$DE//AC\to\dfrac{BD}{BC}=\dfrac{BE}{BA}$
$\to\dfrac{AF}{AC}+\dfrac{BD}{BC}=\dfrac{AE}{AB}+\dfrac{BE}{BA}=\dfrac{AE+BE}{AB}=\dfrac{AB}{AB}=1$
b.Ta có : $\dfrac{S_{AEF}}{S_{BEFC}}=\dfrac{9}{21}$
$\to\dfrac{S_{AEF}}{S_{BEFC}+S_{AEF}}=\dfrac{9}{21+9}$
$\to\dfrac{S_{AEF}}{S_{ABC}}=\dfrac{9}{30}=\dfrac{3}{10}$
Mà $EF//AB\to \widehat{AEF}=\widehat{ABC}\to\Delta AEF\sim\Delta ABC(g.g)$
$\to \dfrac{S_{AEF}}{S_{ABC}}=(\dfrac{AE}{AB})^2$
$\to (\dfrac{AE}{AB})^2=\dfrac{3}{10}$
$\to \dfrac{AE}{AB}=\sqrt{\dfrac{3}{10}}$
$\to \dfrac{AE}{AB}=\dfrac{\sqrt{30}}{10}$
$\to \dfrac{AE}{AB-AE}=\dfrac{\sqrt{30}}{10-\sqrt{30}}$
$\to \dfrac{AE}{BE}=\dfrac{\sqrt{30}}{10-\sqrt{30}}$
Ta có $EF//BC, DE//AC\to\widehat{AEF}=\widehat{EBD},\widehat{BED}=\widehat{EAF}$
$\to\Delta AEF\sim\Delta EBD(g.g)$
$\to\dfrac{P_{AEF}}{P_{BDE}}=\dfrac{AE}{BE}=\dfrac{\sqrt{30}}{10-\sqrt{30}}$
