$ABCD$ is a square and $H$ is an interior point, which divides it for four triangles. If $W$, $X,$ $Y$ and $Z$ are the centroids of triangles $AHD$, $AHB$, $BHC$ and $CHD$ respectively , then what is the ratio between the area of the square $WXYZ$ and the area of the square $ABCD$ ?
Can anyone provide me a hint or a help to go ?
Thank you very much
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Draw the point on $AH$ that's closer to $A$ than $H$ in a 1:2 ratio. Call this $A'$. Repeat for $B$, $C$, and $D$. Show that $A'B'C'D'$ is another square, and that each vertex of $WXYZ$ is the midpoint of an edge of $A'B'C'D'$. Conclude that $WXYZ$ has half the area as $A'B'C'D'$.
Note, by the way, that there's no reason to assume that $H$ is coplanar with $ABCD$. In fact, it feels easier to visualize this problem if you assume it's not, making $ABCDH$ a square prism.
$XW=YZ=\frac{1}{3}BD$, $XY=WZ=\frac{1}{3}AC$ and $XYZW$ is square.
Thus, the ratio is $\frac{2}{9}$ because $S_{ABCD}=\frac{1}{2}AC\cdot BD$ and $$\frac{S_{XYZW}}{S_{ABCD}}=\frac{\frac{1}{3}AC\cdot\frac{1}{3}BD}{\frac{1}{2}AC\cdot BD}=\frac{2}{9}$$
For example, let $M$ is a midpoint of $BH$.
Hence, $$\frac{XY}{AC}=\frac{MY}{MC}=\frac{1}{3}$$