Vì AB//CD
nên \(\dfrac{MA}{MC}=\dfrac{MB}{MD}=\dfrac{AB}{CD}=\dfrac{1}{3}\)
=>\(MC=3\times MA;MD=3\times MB\)
Vì MD=3MB
nên \(S_{AMD}=3\times S_{AMB}\)
Vì MC=3MA
nên \(S_{BMC}=3\times S_{AMB}\)
Do đó: \(S_{AMD}=S_{BMC}\)
Vì AB//CD
nên \(\dfrac{S_{ABD}}{S_{BDC}}=\dfrac{AB}{DC}=\dfrac{1}{3}\)
=>\(S_{DBC}=3\times S_{ABD}\)
Ta có: \(S_{ABD}+S_{BDC}=S_{ABCD}\)
=>\(4\times S_{ABD}=64\)
=>\(S_{ABD}=16\left(cm^2\right)\)