Question 42
Given a triangle $ABC$, take $A'$ on the ray $BA$ (on the opposite side of $A$ to $B$) so that $AA' = BC$, and take $A''$ on the ray $CA$ (on the opposite side of $A$ to $C$) so that $AA'' = BC$. Similarly take $B'$, $B''$ on the rays $CB$, $AB$ respectively with $BB' = BB'' = CA$, and $C'$, $C''$ on the rays $AB$, $CB$. Show that the area of the hexagon $A''A'B''B'C'C''$ is at least 13 times the area of the triangle $ABC$.