The first eight propositions.

To construct an equilateral triangle on a given finite straight line.

Given segment AB, the two circles centred at A and B each passing through the other meet at a point C; the segments CA and CB are radii of those circles, equal to AB, and therefore to each other.

1. Draw circle α centred at A through B.P.3
2. Draw circle β centred at B through A.P.3
3. The circles meet at a point — call it C.Def. 15
4. Draw segments CA and CB.P.1
5. CA = AB (radii of α); CB = BA (radii of β).Def. 15
6. Therefore CA = CB = AB; △ABC is equilateral.CN.1

To place at a given point a straight line equal to a given straight line.

Given a point A and a separate segment BC: construct a segment from A equal in length to BC. The construction is famous for its indirection — Euclid does not permit "carrying" the segment with a compass; he must build a rigid frame from circles and an equilateral triangle.

1. Draw segment AB.P.1
2. Construct equilateral triangle DAB on AB.I.1
3. Produce DA and DB indefinitely.P.2
4. Describe circle centred at B through C; let it meet the extension of DB at G.P.3
5. Describe circle centred at D through G; let it meet the extension of DA at L.P.3
6. BG = BC (radii); DL = DG (radii); DA = DB (equilateral).Def. 15
7. Subtract: AL = DL − DA = DG − DB = BG = BC.CN.3

Given two unequal straight lines, to cut off from the greater a length equal to the less.

Once a length can be transported (I.2) and a circle can be drawn to a given radius (P.3), this becomes immediate: place the shorter length at the endpoint of the longer, swing a circle of that radius, and mark where it meets the longer segment.

1. By I.2, place at A a segment AD equal to C.I.2
2. Describe circle centred at A through D; let it meet AB at E.P.3
3. AE = AD (radii of the same circle).Def. 15
4. AD = C (by construction); therefore AE = C.CN.1

Side–Angle–Side congruence.

If two triangles have two sides equal to two sides respectively, and the angles contained by those equal sides also equal, then the bases are equal, the triangles are equal, and the remaining angles are equal each to each — those opposite the equal sides.

1. Apply △ABC to △DEF so that A falls on D and AB falls along DE.— move
2. Since AB = DE, point B coincides with E.CN.4
3. Since ∠BAC = ∠EDF, ray AC falls along ray DF.— given
4. Since AC = DF, point C coincides with F.CN.4
5. Therefore BC coincides with EF; the triangles coincide.CN.4
6. Coinciding figures are equal in every respect.CN.4

A note on superposition: I.4 is the only proposition in Book I that rests solely on the doctrine of coincidence. Many later mathematicians (including Hilbert) found this awkward — moving a figure presupposes more than the postulates strictly grant. In modern axiomatisations SAS is taken as an axiom rather than a theorem.

In isosceles triangles, the angles at the base are equal to one another.

The "bridge of asses" — the proposition on which medieval students were said to founder. Euclid's proof, unlike the modern symmetry argument, is a tour-de-force of using earlier results: two cuts (I.3) produce two equal triangles outside the original; two applications of SAS (I.4) yield the equality of the base angles by subtraction (CN.3).

1. Let ABC be isosceles with AB = AC; produce both legs.P.2
2. On the extension of AB take any point F; on the extension of AC cut AG = AF.I.3
3. Draw FC and GB.P.1
4. △AFC ≅ △AGB (sides AF, AC; AG, AB; common angle at A).I.4
5. Hence FC = GB, ∠AFC = ∠AGB, ∠ACF = ∠ABG.I.4
6. Subtract AB from AF and AC from AG: BF = CG.CN.3
7. △BFC ≅ △CGB (sides BF, FC; CG, GB; angle at F equal to angle at G).I.4
8. Hence ∠FBC = ∠GCB and ∠BCF = ∠CBG.I.4
9. ∠ABC = ∠ABG − ∠CBG = ∠ACF − ∠BCF = ∠ACB.CN.3

The construction's depth — five postulates, three earlier propositions, two applications of SAS — is what makes this the first proposition where Book I's accumulation becomes visible. It is also where the dependency graph above stops being a tree and becomes a graph.

If two angles of a triangle are equal, the sides opposite them are equal.

The converse of pons asinorum, and Book I's first proof by contradiction. Suppose AB ≠ AC; without loss of generality, AB > AC. Cut off BD = CA on the longer side (I.3) and join DC. By SAS (I.4), △DBC ≅ △ACB — but △DBC sits strictly inside △ACB, so by Common Notion 5 the whole is greater than the part. The contradiction forces AB = AC.

1. Suppose AB ≠ AC. Without loss of generality, AB > AC.hypothesis
2. On BA, cut off BD = CA.I.3
3. Join DC.P.1
4. △DBC ≅ △ACB: DB = AC (by construction), BC common, ∠DBC = ∠ACB (the hypothesis — ∠DBC is the same angle as ∠ABC, since D lies on BA).I.4
5. So △DBC equals △ACB in area.I.4
6. D lies strictly between B and A, so △DBC is a proper part of △ACB.construction
7. The whole is greater than the part: △ACB > △DBC — contradicting the equality just established.CN.5
8. Hence AB > AC is impossible; by symmetry, AB < AC is also impossible; therefore AB = AC. Q.E.D.

Book I's first reductio ad absurdum. The construction stages the figure that the supposed-wrong conclusion forces; the contradiction is then read off the staged figure. This pattern — construct the supposed-wrong figure, derive the contradiction from it — becomes the template for every later reductio in the Elements.

Given two lines on a base meeting at a point, no second pair of equal lines can meet at a different point on the same side.

Book I's second reductio, and a uniqueness lemma whose load-bearing use is I.8 (SSS). Suppose a second apex D on the same side of AB as C with AD = AC and DB = CB. Join CD and inspect the two triangles it produces: pons asinorum (I.5) gives the base angles of each, and a chain of "the whole is greater than the part" (CN 5) forces an angle to be at once greater than and equal to itself. The contradiction shows the supposed second apex cannot exist.

1. Suppose a second apex D on the same side of AB as C, with AD = AC, DB = CB, and D ≠ C.hypothesis
2. Join CD, producing triangles ACD and BCD on the common base CD.P.1
3. In △ACD: AC = AD, so ∠ACD = ∠ADC.I.5
4. D lies inside ∠ACB, so ∠BCD is a part of ∠ACD; hence ∠ACD > ∠BCD.CN.5
5. From (3) and (4): ∠ADC > ∠BCD.
6. D lies inside △ACB, so ∠ADC is a part of ∠BDC; hence ∠BDC > ∠ADC.CN.5
7. From (5) and (6): ∠BDC is much greater than ∠BCD.
8. In △BCD: CB = DB, so ∠BCD = ∠BDC.I.5
9. ∠BDC is at once greater than and equal to ∠BCD. Contradiction.CN.1
10. Therefore no second apex D distinct from C can satisfy the same equalities on the same side. Q.E.D.

A uniqueness lemma. Its work is done in I.8 (SSS congruence): together with the supposition of two triangles equal in all three sides, I.7 forces their apex points to coincide. Heath notes that Euclid's proof covers only one configuration of the two apexes; Proclus and Pappus supplied the cases Euclid omits — the result holds in those cases too, but with the chain of containments adjusted. Common Notion 5 is the lever for the second time in Book I (after I.6); it earns its keep here as the fact that a part is strictly less than its whole, twice over.

If two triangles have the three sides equal, the angles contained by the equal sides are equal.

Side–Side–Side congruence — the third congruence theorem, after I.4 (SAS) and the two reductios I.6, I.7. Its method is superposition, the move first met at I.4: lay the base of one triangle onto the other. Because BC = EF, the base falls exactly — B on E, C on F. But with no angle given, the motion alone cannot fix the apex; that the sides BA, AC cannot fall beside ED, DF is forced by I.7, the apex-uniqueness lemma. I.8 is the payoff I.7 was built for.

1. Apply △ABC to △DEF, placing B on E and BC along EF.— move
2. Since BC = EF, point C coincides with F.— given
3. Suppose BA, AC did not coincide with ED, DF but fell beside them, meeting at a different point G. Then on EF, from its extremities and on the same side, two pairs of equal lines would meet at two distinct points G and D.supposition
4. But I.7 forbids exactly this. So BA, AC coincide with ED, DF.I.7
5. Therefore ∠BAC coincides with ∠EDF, and is equal to it.CN.4

A note on the division of labour: superposition lays the base down, but it is I.7 — not the motion — that closes the apex. At I.4 the included angle carried the apex into place by the motion alone; at I.8 there is no given angle, so apex coincidence must be argued, not moved. I.8 still inherits I.4's unproven superposition assumption for the base placement — leaning on a proven proposition for the apex confines the asserted motion to a smaller part of the proof but does not remove it.