a: ABCD là hình chữ nhật
=>\(S_{ABD}=S_{BCD}=\dfrac{1}{2}\cdot S_{ABCD}\)
ABCD là hình chữ nhật
=>\(S_{ABC}=S_{ADC}=\dfrac{1}{2}\cdot S_{ABCD}\)
b: Vì E là trung điểm của AB
nên \(S_{DEB}=\dfrac{1}{2}\cdot S_{DAB}=\dfrac{1}{4}\cdot S_{ABCD}\)
\(S_{DBC}=\dfrac{1}{2}\cdot S_{ABCD}\)
\(\dfrac{S_{DEB}}{S_{DBC}}=\dfrac{\dfrac{1}{4}\cdot S_{ABCD}}{\dfrac{1}{2}\cdot S_{ABCD}}=\dfrac{1}{4}:\dfrac{1}{2}=\dfrac{2}{4}=\dfrac{1}{2}\)
=>\(S_{DEB}=\dfrac{S_{DBC}}{2}\)
=>\(S_{DEB}< S_{DBC}\)
c: Vì EB//DC
nên \(\dfrac{MB}{MD}=\dfrac{ME}{MC}=\dfrac{EB}{DC}=\dfrac{1}{2}\)
=>MD=2MB
=>\(\dfrac{DM}{DB}=\dfrac{2}{3}\)
=>\(S_{EDM}=\dfrac{2}{3}\cdot S_{EDB}=\dfrac{1}{6}\cdot S_{ABCD}=\dfrac{2010}{6}=335\left(cm^2\right)\)
