Proposition 8 of 48 • Theorem

If two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, then they also have the angles equal which are contained by the equal straight lines.

If two triangles have all three sides equal (SSS - Side-Side-Side), then all their corresponding angles are equal too—the triangles are congruent.

Before You Read

Cut out two triangles from cardboard with all three sides the same length. Can you arrange them so they DON'T overlap perfectly? Try flipping, rotating, whatever you like. Is it possible for two triangles to have identical side lengths but different shapes?

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What Euclid Is Doing

Setup: Triangles ABC and DEF have AB = DE, AC = DF, and BC = EF. All three pairs of sides are equal. We want to prove the triangles are congruent.

Approach: Here's the key insight: imagine picking up triangle ABC and laying it on DEF so the bases match (B on E, C on F). Point A has to land somewhere on the same side as D. Now ask: how far is A from E? It's BA, which equals DE. How far is A from F? It's CA, which equals DF. But wait—D is also exactly DE from E and DF from F. Proposition 7 just told us there can only be ONE point at those distances on a given side. So A must land right on D. The triangles coincide perfectly.

Conclusion: Here's the proof: Place triangle ABC on DEF with B on E and C on F (valid since BC = EF). Point A is now somewhere on the same side of line EF as D. From point E, the distance to A equals BA = DE (given). From point F, the distance to A equals CA = DF (given). So A is a point where: distance from E = DE, and distance from F = DF. But D is also such a point! Proposition 7 says there can only be ONE such point on a given side of EF. Therefore A must coincide with D. Since the triangles coincide completely, ∠BAC = ∠EDF (by Common Notion 4: coincident things are equal). ✓

Key Moves

Given: AB = DE, AC = DF, BC = EF
Superposition: Place triangle ABC onto DEF, aligning B→E and C→F (valid since BC = EF)
Point A lands somewhere on the same side of EF as D
Distance from E to A = BA = DE (given equality)
Distance from F to A = CA = DF (given equality)
D also satisfies: distance from E = DE, distance from F = DF
By Proposition 7: only ONE point can satisfy both distance conditions
Therefore A coincides with D, and ∠BAC = ∠EDF (CN4) ✓

Try It Yourself

Draw a triangle with sides 5 cm, 7 cm, and 9 cm. Now draw another one with the exact same three side lengths. Measure every angle in both. Are they the same? Try it with different sets of lengths—say 4-4-6 or 3-5-7. Can you ever get different angles from the same three side lengths?

Proof Challenge

Available Justifications

Given: Triangles ABC and DEF with AB = DE, AC = DF, BC = EF. Superpose △ABC onto △DEF, aligning B with E and C with F (since BC = EF).

Drag justification

Point A lands on the same side as D. Since BA = ED (given), the distance from E to A equals the distance from E to D.

Drag justification

Since CA = FD (given), the distance from F to A equals the distance from F to D. So A is equidistant from both E and F, just like D.

Drag justification

Two distinct points on the same side of EF cannot both be equidistant from E and F. Therefore A = D.

Drag justification

Since A coincides with D, the triangles coincide entirely and are therefore congruent.

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 8.

✓Lesson Plan (prop-08-lesson-plan.pdf)
✓Student Worksheet (prop-08-worksheet.pdf)
✓Answer Key (prop-08-answer-key.pdf)

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Why It Matters

This is SSS (Side-Side-Side) congruence—the second major congruence theorem. If all three sides of two triangles match, the triangles are identical. This is why three lengths completely determine a triangle's shape.

Modern connection: SSS is the mathematical basis for surveying and navigation. If you can measure three distances, you can determine exact positions. GPS uses a 3D version of this principle (trilateration).

Historical note: Euclid now has two ways to prove triangles congruent: SAS (Prop 4) and SSS (Prop 8). The third common method, ASA, comes later in Proposition 26.

Discussion Questions

How is this different from SAS (Proposition 4)?
If you have a physical triangle and you want to recreate it, which is easier to measure: sides or angles?
Could you have two triangles with three equal sides but different shapes? Why not?

Euclid's Original Proof

Given: Triangles ABC and DEF where AB = DE, AC = DF, and BC = EF (three pairs of equal sides).

Claim: The triangles are congruent, with ∠BAC = ∠EDF.

Proof: Apply △ABC to △DEF so that B falls on E and BC falls along EF.

Since BC = EF, point C falls on F.

Now A falls either: (1) on the same side of EF as D, (2) on the opposite side, or (3) on line EF itself.

Case 1: If A and D are on the same side, then since EA = BA = DE and FA = CA = DF, we have two triangles on the same base with equal pairs of sides meeting at different points—impossible by I.7.

Case 2: If A is on the opposite side from D, the same reasoning applies.

Case 3: If A falls on line EF, then since EA = DE, point A = D.

Therefore A must coincide with D, so △ABC ≅ △DEF and ∠BAC = ∠EDF.

Q.E.D.

What's Next

With SSS congruence in hand, Euclid has a powerful new tool. He puts it to work immediately: Proposition 9 shows how to bisect any angle—cut it exactly in half—using an equilateral triangle and SSS.