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Consider a square. Let $P$ be any point on the base of the square. Connect each of the upper vertices to this point and use these segments as sides to create two more squares (outward). The picture will look like three playing cards being spread out in your hand. Now, use the three centers of the squares and the shared point $P$ to make a quadrilateral. Prove that this quadrilateral will always be a Parallelogram.

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  • $\begingroup$ and your work is.....? $\endgroup$
    – Allie
    Mar 19 '16 at 21:27
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Let $M$ be the linear operator that rotates a vector 45° counterclockwise. One can use $M$ to build the center of a square from its vertices. For instance (see picture): $O=A+{1\over\sqrt2}M(B-A)$. In the same way: $$ R=P+{1\over\sqrt2}M(B-P) \quad\hbox{and}\quad Q=A+{1\over\sqrt2}M(P-A). $$ The midpoint of $QR$ is given by $(Q+R)/2$ and we can express it as follows: $$ {1\over2}(Q+R)= {1\over2}\left(A+{1\over\sqrt2}M(P-A)+P+{1\over\sqrt2}M(B-P)\right)= {1\over2}\left(P+A+{1\over\sqrt2}M(B-A)\right)={1\over2}(P+O). $$ It follows that $QR$ and $PO$ have the same midpoint, thus they form a parallelogram.

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