Giải thích các bước giải:
Ta có $\widehat{AEB}=\widehat{DFC}=90^o$
$AB=CD$ vì $ABCD$ là hình bình hành
$\widehat{BAE}=\widehat{DCF}$ vì $AB//CD$
$\to \Delta ABE=\Delta CDF(g.c.g)$
$\to BE=DF, AE=CF$
$\to CE=AC-AE=AC-CF=AF$
$\to S_{ABDF}=S_{ABF}+S_{ADF}=\dfrac12BE\cdot AF+\dfrac12DF\cdot AF=\dfrac12AF\cdot (BE+DF)$
$\to S_{ABDF}=\dfrac12AF\cdot 2BE$ vì $BE=DF$
$\to S_{ABDF}=AF\cdot BE$
Lại có:
$S_{DEBC}=S_{DEC}+S_{BEC}=\dfrac12DF\cdot CE+\dfrac12BE\cdot CE$
$\to S_{DEBC}=S_{DEC}+S_{BEC}=\dfrac12CE\cdot (DF+BE)$
$\to S_{DEBC}=S_{DEC}+S_{BEC}=\dfrac12AF\cdot 2BE$ vì $BE=DF, CE=AF$
$\to S_{DEBC}=S_{DEC}+S_{BEC}=AF\cdot BE$
$\to S_{ABDF}=S_{DEBC}$
