# Intersection of Altitudes in Hexagon

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!

## 3 Answers

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. 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.

• Oh wow this is a tremendous help, I was not approaching the problem correctly at all. Out of curiosity, would the common chords still intersect if the condition of 3 pairs of equal sides was not met? Does that criteria hold significance to the problem, because it appears as though it does not – Nelly Oct 31 '16 at 4:32
• @Nelly Since the proof does not rely on those given, I imagine that the result will be the same even without the equality of the 3 equal pairs. But I am not that sure and to find out, just draw another picture. – Mick Oct 31 '16 at 4:46
• @Mick: The property does not hold for arbitrary hexagons, so your proof is flawed. (Take property-having $ABCDEF$, with concurrency point $P$, and take $A^\prime$ not on $\overleftrightarrow{AP}$. Then, in $A^\prime BCDEF$, altitudes from $C$ & $E$ still meet at $P$, but the altitude from $A^\prime$ is parallel to the old altitude from $A$, so it does not contain $P$. The new hexagon does not have the property.) It's true that the common chords of the 3 circles concur, but those chords don't always align with the specified altitudes. (That they do here may be equiv to the original problem.). – Blue Oct 31 '16 at 6:37
• @Blue Noted and thanks for the info. – Mick Oct 31 '16 at 13:19

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. $\square$