Menelaus from 3D

Invocation of 3D imagery often serves a powerful tool in proving theorems of plane geometry. Farther down the page there are links to several demonstrations of this idea. On November 28, 2010, Hubert Shutrick had a vision that rendered a one-liner proof of Menelaus' theorem.

Menelaus' theorem

Let three points F, D, and E, lie respectively on the sides AB, BC, and AC of ΔABC or their extensions. Assume just one or all three of the points lie on side extensions. Then the points are collinear iff




= 1

Proof (Hubert Shutrick)

Put up a flagpole AA' orthogonal to the plane and let B' and C' be the points where the orthogonal lines through B and C meet the plane A'DEF. If the heights of the verticals are ha, hb and hc, possibly negative, then similar vertical right triangles give

ha/hb = AF/BF, hb/hc = BD/CD and hc/ha = CE/AE

which, when multiplied together, produce exactly the required identity.

Conversely, assume Menelaus' identity holds. Erect AA' and consider the plane A'EF. Define B' and C' as before, and D' the intersection of B'C' with BC. Then

ha/hb = AF/BF, hb/hc = BD'/CD' and hc/ha = CE/AE.

Comparing the product of the three with Menelaus' identity shows that BD'/CD' = BD/CD, implying D = D' and, subsequently, the collinearity of D with E and F.

2D Problems That Benefit from a 3D Outlook

  1. Four Travellers, Solution
  2. Desargues' Theorem
  3. Soddy Circles and Eppstein's Points
  4. Symmetries in a Triangle
  5. Three Circles and Common Chords
  6. Three Circles and Common Tangents
  7. Three Equal Circles
  8. Menelaus from 3D
  9. Stereographic Projection and Inversion
  10. Stereographic Projection and Radical Axes
  11. Sum of Squares in Equilateral Triangle

Menelaus and Ceva