I've been trying to figure out how CEVA's Theorem can be implemented in solving this problem, but I'm coming up short and cannot make any progress with this problem. The problem states;
A convex hexagon ABCDEF satisfies |AB| = |BC|; |CD| =
|DE|; |EF| = |FA|. Prove that the lines containing the altitudes of the
triangles BCD, DEF, FAB starting respectively at the vertices C, E, A
intersect at a common point.
Any advice or guidance is much appreciated!
Answer
Draw the circles $k_B, \,\, k_D, \,\, k_F$ centered at the vertices $B, \,\, D,\,\, F$ and radii $BA=BC, \,\, DC=DE, \,\, FE=FA$ respectively. Then the altitude line $h_A$ of triangle $ABF$ throguh vertex $A$, the altitude line $h_C$ of triangle $BCD$ throguh vertex $C$ and the altitude line $h_E$ of triangle $DEF$ through vertex $E$ are the radical axes of the three pairs of circles $(k_F, k_B)$, $\,\, (k_B,k_D)$ and $(k_D, k_F)$ respectively. Therefore, by the radical axis theorem, the three radical axes $h_A, h_C$ and $h_F$ intersect at a common point. 
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