Euclid I.24 Proof Why is DFG greater than EGF?

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Proposition 24

If two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base.

Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF respectively, so that AB equals DE, and AC equals DF, and let the angle at A be greater than the angle at D.

I say that the base BC is greater than the base EF. Since the angle BAC is greater than the angle EDF, construct the angle EDG equal to the angle BAC at the point D on the straight line DE. Make DG equal to either of the two straight lines AC or DF. Join EG and FG.

Since AB equals DE, and AC equals DG, the two sides BA and AC equal the two sides ED and DG, respectively, and the angle BAC equals the angle EDG, therefore the base BC equals the base EG.

Again, since DF equals DG, therefore the angle DGF equals the angle DFG. Therefore the angle DFG is greater than the angle EGF. Therefore the angle EFG is much greater than the angle EGF.

Since EFG is a triangle having the angle EFG greater than the angle EGF, and side opposite the greater angle is greater, therefore the side EG is also greater than EF.

But EG equals BC, therefore BC is also greater than EF.

Therefore if two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base.

Why is DFG greater than EGF?

I understand that the proof relies on this to show that GE is larger than EF but it seems to me that the comparability of DFG and EGF is never actually demonstrated. I'm guessing that the presumption is that whenever Triangle DGF is superimposed on Triangle EGF it looks like Figure 2. But how do you know it looks like Figure 2? What if it looks like Figure 3? Although DFG is still greater than EGF there, it's also greater than EFG and the proof seems to rely on DFG being smaller than EFG but greater than EGF. It seems like the proof isn't universal, though the proposition certainly is (presuming you're working on a plane). Did I miss something in the proof?

Thanks folks!

EDIT: Figs 2 & 3 are attempts to show what might happen if Triangles DGF and EGF are put on the same plane. The main question is where does the statement that this proof seems to rely on (angle DGF is greater than angle EGF) actually come from? I don't recall seeing anything suggestive of it in the previous 23 propositions.

Figure 1: a triangular prism. Figure 2: an isoceles triangle superimposed upon a scalene triangle. DGF is the isoceles triangle. EGF is the scalene triangle and FE is outside of the triangle. Figure 3: EGF is contained inside DGF

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Pay attention to this line:

Since the angle BAC is greater than the angle EDF, construct the angle EDG equal to the angle BAC at the point D on the straight line DE.

Or, put it a little bit differently. Let $G$ be on the same half plane as $F$ (divided by the line $DE$), such that $\widehat{EDG} = \widehat{BAC}$, and $DG = DF ( = AC)$.

So, basically, your drawing in Pic 3 is not what the author is referring to. Fig 2 is also incorrect. And $\widehat{DFG}>\widehat{EGF}$ is because of the 2 following facts:

  • $\widehat{DFG} = \widehat{DGF}$ (since $DF = DG$, it's the way how we construct $G$)
  • $\widehat{DGF}>\widehat{EGF}$