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I have difficulties in understanding the 5th proposition of Euclid's elements in the first book:

Proposition I.5. The base angles in an isosceles triangle are congruent. If the sides of an isosceles triangle are extended beyond the base, the angles formed under the base are congruent.

Proof.

Let $\triangle ABC$ be isosceles with $AB \cong AC$. Extending $AB$ and $AC$ past the base $BC$, we can invoke Prop. I.3 to choose $D$ on $AB$ and $E$ on $AC$ so that $AD \cong AE$. Form segments $DC$ and $EB$. We have SAS (I.4) for $\triangle ADC \cong \triangle AEB$, since $\angle CAD$ is common. It follows that $DC \cong EB$, $\angle ACD \cong \angle ABE$, and $\angle ADC \cong \angle AEB$.

Since the whole $AD \cong AE$, and the part $AB \cong AC$, [Common Notion 3] implies $BD \cong CE$. Since $\angle BDC \cong \angle BEC$, and $BC$ is common, we have SAS for $\triangle BDC \cong \triangle BEC$.

From there, $\angle CBD \cong \angle BCE$, which proves the second assertion of the proposition. We also have $\angle BCD \cong \angle CBE$. Since $\angle ABE \cong \angle ACD$, [Common Notion 3] gives us $\angle ABC \cong \angle ACB$, which proves the result. $\square$

enter image description here

I don't undestand this part:

Since $\angle BDC \cong \angle BEC$, and $BC$ is common, we have SAS for $\triangle BDC \cong \triangle BEC$.

Proposition I.4 which is SAS, states something different:

Proposition I.4 (SAS). If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also equal.

Proposition I.5 in the text I'm reading is considering only $BC$ and not two equal sides and angle comprised between them.

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  • $\begingroup$ See text with comment. $\endgroup$ – Mauro ALLEGRANZA Jul 31 '19 at 11:18
  • $\begingroup$ in the image in that link there are no points C and E when it refers to CBD equals the angle BCE. $\endgroup$ – Michaelangelo Meucci Aug 1 '19 at 8:09
  • $\begingroup$ Imagine $C$ further on line $ABF$ and the same for $E$. $\endgroup$ – Mauro ALLEGRANZA Aug 1 '19 at 10:01
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In triangles $BDC$ and $CEB$ you know that:

1) $CD\cong BE$;

2) $\angle BDC\cong\angle CEB$;

3) $BD\cong CE$;

4) $BC$ is in common.

Hence $BDC\cong CEB$ either by SAS (1, 2, 3) or by SSS (1, 3, 4).

EDIT.

Euclid was forced to use SAS because he proved SSS only in later Proposition (I.8). He added that $BC$ is in common probably to make clear to the reader that he was referring to Prop. I.4 (see Heath's comment quoted in Mauro ALLEGRANZA's comment below).

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  • $\begingroup$ @MauroALLEGRANZA Yes, and it's the same chain of reasoning used in the question. The only difference is that the text in the question also mentions that triangles $BDC$ and $CEB$ have $BC$ in common, which is irrelevant. $\endgroup$ – Intelligenti pauca Jul 31 '19 at 13:39
  • $\begingroup$ Presumably, that was Euclid's text ... $\endgroup$ – Mauro ALLEGRANZA Jul 31 '19 at 13:43
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    $\begingroup$ @MauroALLEGRANZA I stand corrected: I've just checked Fitzpatrick's translation of Heiberg's text and the common side is there! $\endgroup$ – Intelligenti pauca Jul 31 '19 at 14:10
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    $\begingroup$ See Dover ed page 252 (Heath comment) : "the base BC is common to them, i.e., apparently, common to the angles. Simson wrote "and the base BC is common to the two triangles BFC, CGB"; Todhunter left out these words as being of no use and tending to perplex a beginner. But Euclid evidently chose to quote the conclusion of I. 4 exactly; the first phrase of that conclusion is that the bases (of the two triangles) are equal, and, as the equal bases are here the same base, Euclid naturally substitutes the word "common" for "equal." " $\endgroup$ – Mauro ALLEGRANZA Jul 31 '19 at 14:17
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    $\begingroup$ But the "textual" issue is that Euclid's text is a copy of a copy of a copy... There are still discussions about the possibility that some of the five axioms are later interpolation; thus, I think that it is quite impossible to ascertain if that little statement was due to Euclid's "sloppiness" or is some later intepolation due to some over-zelous copyst. $\endgroup$ – Mauro ALLEGRANZA Jul 31 '19 at 14:43

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