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Euclid, Proposition I.4

Proposition I.4 presents Euclid’s proof of what is called the “Side-Angle-Side Theorem” in modern schools. The question students often ask is in Classical Geometry, when studying proposition 1.4 is, “How can we just assume the two triangles with two equal sides and a common angle between them?”

That does seem to come out of nowhere.

These two imagined triangles can be produced using the Elements and Propositions I.1-3. See image above. To summarize:

  1. Transfer line AB to point D. (Prop 1.2)
  2. Transfer a line equal to AC to point D. (Prop 1.2)
  3. Transfer a line equal to BC to point I. (Prop 1.2)
  4. Draw circle from point D at distance K. (Post. 1.3)
  5. Draw circle from point I at distance P. (Post 1.3)
  6. Point Q must exist at the point where circles IP and DK intersect.

Those who have studied propositions I.1-3 should be able to see how this was done. Follow the letters in alphabetical order. I haven’t formally proven this yet, but I think it can be proven. The ultimate question is how to place point Q, which I believe can be done at the point where circle IP and circle DK intersect.

I’m assuming that the reason why Euclid (?) attempts to demonstrate this theorem indirectly is to show that if point Q is placed anywhere other than where the two circles instersect, no triangle composed of straight lines would exist.

Anyway, this is for the Quadrivi-ites.

Mr. William C. Michael, O.P.
Classical Liberal Arts Academy

4 Comments

  1. Jake Le Master Jake Le Master February 15, 2024

    I assume you mean AC to point D in step 2 (creating line DK).

    At a glance, this appears to be more elegant than Euclid’s superposition of “lifting” one triangle and placing it on the other.

    Drawing a line equal to AB off point D, and a line equal to BC off point I gets us a way to have AB and DI to coincide, and BC and IQ to coincide since they are radii of circles described by points D and I, respectively.

    After that though, I can’t seem to prove that DK (equal to AC) coincides with DQ.

    • williamcmichael williamcmichael Post author | February 15, 2024

      Thanks for the typo, I fixed step 2 to read at point D.

      Point Q is located at the intersection of the circles DK and IP. Therefore DQ must equal DK, which is equal to AC.

      • Jake Le Master Jake Le Master February 15, 2024

        That’s what we want, but I’m not sure if we can assume they intersect at Q. It seems to beg what we’re trying to prove — that DK and DQ are radii of the same circle, and are therefore equal.

        I was initially thinking there should be two cases in which a reductio is done. One where AC is assumed to be shorter than DQ, and one in which DQ is assumed to be shorter than AC. For example, if AC is assumed to be shorter than DQ, cut off a new line [DM] from DQ equal to AC. That new angle DIM will then be smaller than DIQ. — which can’t be. Except I’m not sure this works, because it then assumes the new line IM is equal to IQ (and thus BC), which I don’t think can be assumed.

        • williamcmichael williamcmichael Post author | February 15, 2024

          “Q” is placed at the point where the circles intersect. We do the same thing in Prop. 1.1 when we place a point at the place where the circles meet, which becomes the third point of the equilateral triangle. If we can do that in Prop. 1.1, we should be able do the same thing in 1.4.

          This is not the proof, however, only the construction of the two triangles. The proof Euclid provides shows that if Q is placed anywhere other than that point, there could not be a triangle.

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