Giải thích các bước giải:
Ta có $BD\perp BC\to\widehat{DBC}=\widehat{DAB}=90^o$
Mà $\widehat{ABD}=\widehat{BDC}(AB//CD)$
$\to\Delta ABD\sim\Delta BDC(g.g)$
$\to\dfrac{BD}{DC}=\dfrac{AB}{BD}$
$\to DC=\dfrac{BD^2}{AB}$
Ta có:
$AB//CD\to \dfrac{EA}{EC}=\dfrac{AB}{DC}=\dfrac{AB^2}{BD^2}$
$\to \dfrac{EA}{EA+EC}=\dfrac{AB^2}{AB^2+BD^2}$
$\to\dfrac{EA}{AC}=\dfrac{AB^2}{AB^2+BD^2}$
$\to\dfrac{S_{AED}}{S_{ADC}}=\dfrac{AB^2}{AB^2+BD^2}$
$\to S_{AED}=\dfrac{AB^2}{AB^2+BD^2}\cdot S_{ADC}$
$\to S_{AED}=\dfrac{AB^2}{AB^2+BD^2}\cdot \dfrac12\cdot AD\cdot CD$