Given triangle $ABC$, let $A’$ be the point 1/3 of the way from $B$ to $C$, as shown. Similarly, $B’$ is the point 1/3 of the way from $C$ to $A$, and $C’$ is the point 1/3 of the way from $A$ to $B$. In this way, we have constructed a new triangle starting with an arbitrary triangle. Now apply the same procedure to triangle $A’B’C’$, thereby creating triangle $A’’B’’C’’$. Show that the sides of triangle $A’’B’’C’’$ are parallel to the (appropriate) sides of triangle $ABC$. What fraction of the area of triangle $ABC$ is the area of triangle $A’’B’’C’’$?

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