Intersection of Altitudes in Hexagon

Question

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 1

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.

Answer 2

Let P, Q, R be the feet of the said altitudes (in blue).

Then, red, purple, and green circles can be formed. Those blue lines happen to be the common chords. They concur at the common point X, the radical center.

Answer 3

If Ceva's Theorem isn't required, then I recommend Carnot's Theorem (not to be confused with Carnot's Theorem).

Let the pairs of edges meeting $B$, $D$, $F$ have respective lengths $b$, $d$, $f$, and let the altitudes from $A$, $C$, $E$ have lengths $a$, $c$, $e$. With $A^\prime$, $C^\prime$, $E^\prime$ the feet of those respective altitudes, we have ...

$$\begin{align} |\overline{BA^\prime}|^2 + |\overline{DC^\prime}|^2 + |\overline{FE^\prime}|^2 &= \left(\;b^2 - a^2\;\right) + \left(\;d^2 - c^2\;\right) + \left(\;f^2 - e^2 \;\right) \\[6pt] &= \left(\;f^2 - a^2\;\right) + \left(\;b^2 - c^2\;\right) + \left(\;d^2 - e^2 \;\right) \\[6pt] &= |\overline{FA^\prime}|^2 + |\overline{BC^\prime}|^2 + |\overline{DE^\prime}|^2 \end{align}$$ By Carnot, the perpendiculars to the sides of $\triangle BDF$ at points $A^\prime$, $C^\prime$, $E^\prime$ are concurrent. $\square$