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In the given figure,ABCD is a quadrilateral and E,F G and H are respectively the mid-points of its sides. Prove that the area of the parallelogram EFGH formed by joining the mid-points of the sides of the quadrilateral is half the area of the quadrilateral.

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The question can be easily proved if the ABCD is a parallelogram but as the ABCD is a quadrilateral it is being difficult. How to go about it?

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$EF$ is the basis media of triangle $ABC$, and then $A(BEF)=1/4A(ABC)$. Idem $A(GHD)=1/4A(ACD)$.

Now, note that $A(ACD)+A(ABC)=A(ABCD)\to A(BEF)+A(GHD)=1/4A(ABCD)[*_1]$.

Idem, $ A(HEA)+A(GFC)=1/4A(ABCD)[*_2]$.

Adding $ [*_1] +[*_2]$ the problem is solved

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  • $\begingroup$ Excellent !! @vvnitram $\endgroup$ – Vinay5forPrime Mar 20 '16 at 15:03
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HINT.-The shortest way could be, I think, the vector viewpoint.

Let $A,B,C,D$ the vertices of the quadrilateral; $A',B',C',D'$ the midpoints of segments $AB,BC,CD,DA$ respectively. We have

Area $\triangle ABD=\frac 12||\vec{CB}\text{x}\vec{CD}||$ and Area $\triangle DBC=\frac 12||\vec{AB}\text{x}\vec{AD}||$ Hence the area of the quadrilateral is the sum of these two areas.

Similarly, area $\triangle D'A'C'=\frac 12||\vec{D'A'}\text{x}\vec{D'C'}||$ and Area $\triangle C'A'B'=\frac 12||\vec{B'C'}\text{x}\vec{B'A'}||$

Now it is clear because $\vec{OA'}=\frac{\vec{OA}+\vec{OB}}{2}$ et cetera.

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  • $\begingroup$ Rather good and abstract approach! Never thought of it in this manner.@Piquito $\endgroup$ – Vinay5forPrime Mar 22 '16 at 7:03

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