prove that the area of the figure obtained by joining the mid points of the sides of a quadrilateral is half than that of the quadrilateral
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Now, note that A(ACD)+A(ABC)=A(ABCD)→A(BEF)+A(GHD)=1/4A(ABCD)[∗1]A(ACD)+A(ABC)=A(ABCD)→A(BEF)+A(GHD)=1/4A(ABCD)[∗1].
Idem, A(HEA)+A(GFC)=1/4A(ABCD)[∗2]A(HEA)+A(GFC)=1/4A(ABCD)[∗2].
Adding [∗1]+[∗2][∗1]+[∗2] the problem is solved