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The Saccheri Quadrilateral

We’ve been exploring non-Euclidean geometries lately, and the Saccheri Quadrilateral plays a pivotal role in this particular mathematical history.

The Saccheri quadrilateral is a biperpendicular quadrilateral with two congruent legs.  It’s an object that is “obviously” a rectangle in Euclidean geometry, but proving that without the aid of the parallel postulate turns out to be rather tricky.

In fact, just proving that the measure of the green angle is less than the measure of the orange angle is pretty tough!

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  1. March 25, 2012 at 10:13 pm

    You mentioned on Twitter that you’re looking for help with a proof of this. Unfortunately, there isn’t one—because it isn’t true! (At least if I’m understanding you correctly) The “weak” exterior angle theorem—that an exterior angle of a triangle is greater than either of its remote interior angles (Elements I.16)—isn’t valid on a sphere. Consider a triangle with a 100º interior angle at the north pole and two right interior angles at the equator. The exterior angles at the equator are also right, and so less than the 100º interior angle at the pole.

    The same thing happens in your Saccheri quadrilateral. Make a Saccheri quadrilateral with a 100º pole angle, then lower its “top” side closer and closer to the equator. Its summit angles then get as close to 90º as you please (since the “top” side acts more and more like the equator; or if you prefer, its area goes to 0, and so its angle sum goes to a Euclidean 360º). And presto—an example of a Saccheri quadrilateral on a sphere whose exterior green angle is greater than its summit orange angles.

    A final note: Euclid’s proof of the weak exterior angle theorem fails on a sphere because he uses his second postulate—that any straight line can be extend indefinitely—to determine a point outside of the given triangle. But on a sphere, extending this line in this fashion doesn’t necessarily give a point outside of the triangle, because the line could wrap all of the way around the sphere and back into the triangle!

    I hope that’s useful. Happy non-Euclideaning!

    • March 25, 2012 at 10:31 pm

      So I didn’t catch onto your question quite right, but the thrust of my argument still holds: if you make the pole angle obtuse, the green angle can get as close to equalling that obtuse angle as you please, and the orange summit angles can get as close to right as you please by pushing the quadrilateral’s top toward the equator.

  2. March 25, 2012 at 10:31 pm


    Thanks for the well-articulated (and cited!) reply. I think I understand your argument, but I don’t think it speaks exactly to what I’m trying to show (probably because there is some information and context that hasn’t been provided in this particular problem).

    I’m not looking to prove the exterior angle theorem here. This is part of the proof that, in a Saccheri quadrilateral with unequal legs, the longer side is opposite the larger angle.

    The green angle is part of the original biperpendicular quadrilateral with distinct legs, and the leg to the green angle is the longer leg. We’re trying to prove that the green angle is the smaller of the two original summit angles. This seems to rely on the fact that the green angle is indeed less than the orange angle, which is the summit angle of a Saccheri quadrilateral constructed inside the original biperpendicular quadrilateral.

    Hopefully that clears things up. But I’m still not sure (a) how to prove it, or (b) if it’s true!

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