Ta có: \(BM=\dfrac{1}{2}AB\Rightarrow S_{BDM}=\dfrac{1}{2}S_{ABD}\) (cùng đường cao)
Tương tự, ta có: \(BN=\dfrac{1}{2}BC\Rightarrow S_{BDN}=\dfrac{1}{2}S_{BCD}\) (cùng đường cao)
Suy ra: \(S_{BDM}+S_{BDN}=\dfrac{1}{2}S_{ABD}+\dfrac{1}{2}S_{BCD}=\dfrac{1}{2}\left(S_{ABD}+S_{BCD}\right)\) hay \(S_{ABCD}=2S_{DMBN}\)
Từ D hạ DH vuông AB
SABCD=DH.AB=S
Mà SDMB=1/2DH.MB
=1/2DH.AB
=1/4S
Từ D hạ DK vuông BC, ta có :
SABCD=DK.BC=S
Mà SDMN=1/2DK.BN=1/2DK.1/2BC
=1/4DK.BC=1/4S
Vậy SDMBN=SDMB+SDBN
=1/4S+1/4S=2/4S
=1/2S
